How do you find the indefinite integral of int (-4/x^3-8/x^5) dx?

Nov 27, 2016

The answer is $= \frac{2}{x} ^ 2 + \frac{2}{x} ^ 4 + C$

Explanation:

We use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

So, $\int \left(- \frac{4}{x} ^ 3 - \frac{8}{x} ^ 5\right) \mathrm{dx} = \int \left(- 4 {x}^{- 3} - 8 {x}^{- 5}\right) \mathrm{dx}$

$= - 4 \cdot {x}^{- 3 + 1} / \left(- 3 + 1\right) - 8 \cdot {x}^{- 5 + 1} / \left(- 5 + 1\right) + C$

$= \frac{2}{x} ^ 2 + \frac{2}{x} ^ 4 + C$