# How do you find the antiderivative of 1/x^2?

$\int \frac{1}{x} ^ 2 \mathrm{dx} = \int {x}^{-} 2 \mathrm{dx} = {x}^{-} \frac{1}{-} 1 + c = - \frac{1}{x} + c$
where you used the fact that: $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + c$