# How do you find the indefinite integral of int (-4x^-4-20x^-5) dx?

##### 1 Answer
Sep 9, 2016

$= \frac{1}{3} {x}^{-} 4 \left(4 x + 15\right) + C$.

#### Explanation:

We use the Standard Form $: \int {t}^{n} \mathrm{dt} = {t}^{n + 1} / \left(n + 1\right) , n \ne - 1$.

Hence, $\int \left(- 4 {x}^{-} 4 - 20 {x}^{-} 5\right) \mathrm{dx}$

$\int - 4 {x}^{-} 4 \mathrm{dx} - \int 20 {x}^{-} 5$

$- 4 \int {x}^{-} 4 \mathrm{dx} - 20 \int {x}^{-} 5 \mathrm{dx}$

$= - 4 \frac{{x}^{- 4 + 1}}{- 4 + 1} - 20 \frac{{x}^{- 5 + 1}}{- 5 + 1}$

$\frac{4}{3} {x}^{-} 3 + 5 {x}^{-} 4$

$= \frac{1}{3} {x}^{-} 4 \left(4 x + 15\right) + C$.