# What is the integral of int (x)/(x+1)^3 dx?

Feb 4, 2016

$\frac{- 1}{x + 1} + \frac{1}{2 {\left(x + 1\right)}^{2}} + C = \frac{- 2 x - 1}{2 {\left(x + 1\right)}^{2}} + C$

#### Explanation:

$\int \frac{x}{x + 1} ^ 3 \mathrm{dx} = \int \frac{x + 1 - 1}{x + 1} ^ 3 \mathrm{dx}$

$= \int \left(\frac{x + 1}{x + 1} ^ 3 - \frac{1}{x + 1} ^ 3\right) \mathrm{dx}$

$= \int \left({\left(x + 1\right)}^{-} 2 - {\left(x + 1\right)}^{-} 3\right) \mathrm{dx}$

Let $u = x + 1$ if you want to include those steps.

$= \frac{- 1}{x + 1} + \frac{1}{2 {\left(x + 1\right)}^{2}} + C$

Do the algebra to get a single ration if you wish.