# How do you integrate int 1/x^4dx?

$- \frac{1}{3 {x}^{3}} + C$
We will have to know the rule $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$.
Rewriting the integral gives $\int \frac{1}{x} ^ 4 \mathrm{dx} = \int {x}^{-} 4 \mathrm{dx}$
Now using the rule: $\int {x}^{-} 4 \mathrm{dx} = {x}^{- 4 + 1} / \left(- 4 + 1\right) = {x}^{-} \frac{3}{- 3} = - \frac{1}{3 {x}^{3}} + C$