How do you integrate #int 1/x^3dx#?

Question

Christopher Agresta · Accepted Answer

Rewrite as #intx^-3dx# and take the anti-derivative.

We first need to recognize that #1/x^3# is equivalent to #x^-3#. After we reach that point, solving the issue becomes fairly easy. However, exercise caution because we are working with a negative exponent, which means that the power will decrease as we add one to the power when taking the anti-derivative. This also means that our constant must be negative because the integral's x term is positive. After taking the anti-derivative, we obtain: #-1/2*x^-2 +C# This is equivalent to #-1/(2x^2) +C# As usual, you can check this answer by taking the derivative, which gives you #x^-3#.

Christopher Agresta · Answer

undefined To integrate $ \int \frac{1}{x^3} \, dx $, you can use the power rule for integration, which states that $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $ for any real number $ n \neq -1 $, where $ C $ is the constant of integration. Applying this rule:

$$
\int \frac{1}{x^3} \, dx = \frac{1}{-2}x^{-2} + C
$$

So, the integral of $ \frac{1}{x^3} $ with respect to $ x $ is $ -\frac{1}{2x^2} + C $, where $ C $ is the constant of integration.

Cameron Alexander · Answer

undefined To integrate $ \int \frac{1}{x^3} \, dx $, use the power rule for integration:

$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $$

Applying this rule to $ \int \frac{1}{x^3} \, dx $, where $ n = -3 $, gives:

$$ \int \frac{1}{x^3} \, dx = \frac{x^{-3+1}}{-3+1} + C $$

$$ = \frac{x^{-2}}{-2} + C $$

$$ = -\frac{1}{2x^2} + C $$