# What is the antiderivative of 1/x^3?

$- \frac{1}{2 {x}^{2}} + C$
$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$.
So $\int \frac{1}{x} ^ 3 \mathrm{dx} = \int {x}^{- 3} \mathrm{dx} = - \frac{1}{2} {x}^{- 2} + C = - \frac{1}{2 {x}^{2}} + C$