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

Oct 21, 2016

$- \frac{1}{x} ^ 4 + C$

#### Explanation:

First pull out the constant using $\int a f \left(x\right) \mathrm{dx} = a \int f \left(x\right) \mathrm{dx}$:

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

Now, use the rule for integrating functions of $x$ such as these: $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

Thus

$4 \int {x}^{-} 5 \mathrm{dx} = 4 \left({x}^{- 5 + 1} / \left(- 5 + 1\right)\right) + C = 4 \left(\frac{{x}^{-} 4}{- 4}\right) + C = - \frac{1}{x} ^ 4 + C$