What is the antiderivative of 1/ x^2?

Sep 10, 2016

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

Explanation:

Use the Power Rule

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \text{ , where } n \ne - 1$

here, $n = - 2$

$\int {x}^{- 2} \mathrm{dx}$

$= {x}^{- 2 + 1} / \left(- 2 + 1\right) + C$

$= {x}^{- 1} / \left(- 1\right) + C$

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