How do you find the antiderivative of #2/z^2#?

Answer 1

Represent the fraction as a negative exponent and use the reverse power rule.

#2/z^2# can be rewritten as #2z^-2# So our integral becomes: #int 2z^-2 dz# we can now pull out the #2# from the integral as it is a constant. #2int z^-2 dz# The reverse power rule states #int x^n dx = x^(n+1)/(n+1) + c# Thus #2int z^-2 dz = 2*[z^(-2+1)/(-2+1)] = 2*[z^-1/-1] = 2*[-1/z]# Simplifying this we get: #-2/z + c# We add a #+c# at the end because the integral is indefinite.
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Answer 2

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To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

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Applying this rule to ( \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

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Applying this rule to ( \frac{2}{z^2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ),To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( nTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n =To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

\To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \beginTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{alignedTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \intTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

\To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \intTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dzTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dzTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &=To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz =To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \intTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dzTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dzTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz =To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &=To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \intTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdotTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dzTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

UsingTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using theTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the powerTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power ruleTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} +To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + CTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, whereTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( nTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &=To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n =To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} +To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + CTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} ,To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \endTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

SoTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, theTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivativeTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative ofTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} +To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + CTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C =To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} )To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + CTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, theTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z )To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) isTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} +To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} )To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) isTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ),To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), whereTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where (To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\fracTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C )To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) isTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is theTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{zTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constantTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z}To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constant ofTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z} +To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constant of integrationTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z} + CTo find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constant of integration.To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z} + C \To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constant of integration.To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z} + C ),To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ), where ( n = -2 ):

[ \begin{aligned} \int \frac{2}{z^2} , dz &= 2 \int z^{-2} , dz \ &= 2 \cdot \frac{z^{-2+1}}{-2+1} + C \ &= -\frac{2}{z} + C. \end{aligned} ]

So, the antiderivative of ( \frac{2}{z^2} ) with respect to ( z ) is ( -\frac{2}{z} + C ), where ( C ) is the constant of integration.To find the antiderivative of ( \frac{2}{z^2} ), you can use the power rule for integration, which states that the antiderivative of ( z^n ) with respect to ( z ) is ( \frac{z^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying this rule to ( \frac{2}{z^2} ):

[ \int \frac{2}{z^2} , dz = \int 2z^{-2} , dz = 2 \int z^{-2} , dz ]

Using the power rule, where ( n = -2 ):

[ \int z^{-2} , dz = \frac{z^{-2+1}}{-2+1} + C = \frac{-1}{z} + C ]

Therefore, the antiderivative of ( \frac{2}{z^2} ) is ( -\frac{2}{z} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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