# How do you integrate x/(x+1)^3?

Jul 13, 2018

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

#### Explanation:

Perform this integral by substitution

Let $u = x + 1$, $\implies 0$, $\mathrm{du} = \mathrm{dx}$

The integral is

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

$= \int \left({u}^{-} 2 - {u}^{-} 3\right) \mathrm{du}$

$= - \frac{1}{u} + \frac{1}{2 {u}^{2}}$

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

Jul 13, 2018

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

#### Explanation:

Here,

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

Subst . color(blue)(x+1=u=>x=u-1=>dx=du

So,

$I = \int \frac{u - 1}{u} ^ 3 \mathrm{du}$

$= \int \left[\frac{1}{u} ^ 2 - \frac{1}{u} ^ 3\right] \mathrm{du}$

$= \int \left[{u}^{-} 2 - {u}^{-} 3\right] \mathrm{du}$

$= {u}^{- 2 + 1} / \left(- 2 + 1\right) - {u}^{- 3 + 1} / \left(- 3 + 1\right) + c$

$= {u}^{-} \frac{1}{- 1} - {u}^{-} \frac{2}{- 2} + c$

$= - \frac{1}{u} + \frac{1}{2 {u}^{2}} + c$

$= \frac{1}{2 {u}^{2}} - \frac{1}{u} + c$

$= \frac{1 - 2 u}{2 {u}^{2}} + c$

Subst. back color(blue)(u=x+1

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

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