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How do you integrate #x^-1#?

Answer 1

Emily Ammerman

#intx^-1dx=int1/xdx=ln|x|+C#.

The Standard Form #int x^ndx=x^(n+1)/(n+1)# is not applicable here, as for #n=-1, (n+1)=0;" &, we can't divide by "#0#.

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Answer 2

Ezekiel Alexander

To integrate ( x^{-1} ), you can use the formula for integrating powers of ( x ), which is:

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

Applying this formula to ( x^{-1} ), where ( n = -1 ), we get:

[ \int x^{-1} , dx = \frac{x^{-1+1}}{-1+1} + C = \ln|x| + C ]

So, the integral of ( x^{-1} ) is ( \ln|x| + C ), where ( C ) is the constant of integration.

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Answer 3

Ezekiel Alexander

The integral of ( x^{-1} ) is equal to the natural logarithm of the absolute value of x, plus a constant of integration:

[ \int x^{-1} , dx = \ln(|x|) + C ]

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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.