# Question #a20af

Apr 2, 2017

$\int \frac{x {e}^{x}}{x + 1} ^ 2 \mathrm{dx} = {e}^{x} / \left(x + 1\right) + C$

#### Explanation:

$I = \int \frac{x {e}^{x}}{x + 1} ^ 2 \mathrm{dx}$

Performing integration by parts, let:

$u = {e}^{x} \text{ "=>" } \mathrm{du} = {e}^{x} \textcolor{w h i t e}{.} \mathrm{dx}$

$\mathrm{dv} = \frac{x}{x + 1} ^ 2 \mathrm{dx} \text{ "=>" } v = \ln \left\mid x + 1 \right\mid + \frac{1}{x + 1}$

Note that you can find $\int \frac{x}{x + 1} ^ 2 \mathrm{dx}$ by letting $t = x + 1$, showing that $\int \frac{x}{x + 1} ^ 2 \mathrm{dx} = \int \frac{t - 1}{t} ^ 2 \mathrm{dt} = \ln \left\mid t \right\mid + \frac{1}{t}$.

Then:

$I = {e}^{x} \ln \left\mid x + 1 \right\mid + {e}^{x} / \left(x + 1\right) - \int {e}^{x} \ln \left\mid x + 1 \right\mid \mathrm{dx} - \int {e}^{x} / \left(x + 1\right) \mathrm{dx}$

Now perform integration by parts on $\int {e}^{x} / \left(x + 1\right) \mathrm{dx}$. Let:

$u = {e}^{x} \text{ "=>" } \mathrm{du} = {e}^{x} \textcolor{w h i t e}{.} \mathrm{dx}$

$\mathrm{dv} = \frac{1}{x + 1} \mathrm{dx} \text{ "=>" } v = \ln \left\mid x + 1 \right\mid$

So:

$I = {e}^{x} \ln \left\mid x + 1 \right\mid + {e}^{x} / \left(x + 1\right) - \int {e}^{x} \ln \left\mid x + 1 \right\mid \mathrm{dx} - \left({e}^{x} \ln \left\mid x + 1 \right\mid - \int {e}^{x} \ln \left\mid x + 1 \right\mid \mathrm{dx}\right)$

Many of these terms cancel, leaving just:

$I = {e}^{x} / \left(x + 1\right)$

Attaching the constant of integration:

$I = \int \frac{x {e}^{x}}{x + 1} ^ 2 \mathrm{dx} = {e}^{x} / \left(x + 1\right) + C$