# What is the antiderivative of (1/(x^2))?

$- \frac{1}{x} + C$
$\frac{d}{\mathrm{dx}} \left({x}^{r} + C\right) = r {x}^{r - 1}$ (for constant $C$)
So, for all $s \ne 0$, the antiderivative of ${x}^{s}$ is ${x}^{s + 1} / \left(s + 1\right) + C$.
$\frac{1}{x} ^ 2 = {x}^{-} 2$ so its antiderivative is ${x}^{- 2 + 1} / \left(- 2 + 1\right) = {x}^{-} \frac{1}{-} 1 = - \frac{1}{x}$ don't forget the $+ C$.