x e^ 2 x (1+2 x)^2 d x=( where C is a constant of integration.)

MCQ

$\int \frac{x e^{2 x}}{(1+2 x)^2} d x=($ where $C$ is a constant of integration.$)$

A
$\frac{e^{2 x}}{1+2 x}+C$
B
$\frac{e^{2 x}}{2(1+2 x)}+C$
C
$\frac{4 e^{2 x}}{1+2 x}=C$
✓
$\frac{e^{2 x}}{4(1+2 x)}+C$

✓

Answer

Correct option: D.

$\frac{e^{2 x}}{4(1+2 x)}+C$

Using partial fraction, we have
$ \frac{x}{(1+2 x)^2}=\frac{A}{1+2 x}+\frac{B}{(1+2 x)^2}$
$\Rightarrow x=A(1+2 x)+B$
For $x=-1 / 2, B=-1 / 2$
For $x=0, A=1 / 2$
$\therefore \frac{x}{(1+2 x)^2}=\frac{1}{2(1+2 x)}-\frac{1}{2(1+2 x)^2}$
$\Rightarrow \int \frac{x e^{2 x}}{(1+2 x)^2} d x=\int e^{2 x}\left[\frac{1}{2(1+2 x)}-\frac{1}{2(1+2 x)^2}\right] d x$
$=\frac{1}{2} \int \frac{e^{2 x}}{1+2 x} d x-\frac{1}{2} \int \frac{e^{2 x}}{(1+2 x)^2} d x$
$=\frac{1}{2}\left[\frac{1}{1+2 x} \int e^{2 x} d x-\int\left[\frac{d}{d x}\left(\frac{1}{1+2 x}\right) \int e^{2 x} d x\right] d x\right.\left.-\int \frac{e^{2 x}}{(1+2 x)^2} d x\right]$
$=\frac{1}{2}\left[\frac{e^{2 x}}{2(1+2 x)}+\int \frac{e^{2 x}}{(1+2 x)^2} d x-\int \frac{e^{2 x}}{(1+2 x)^2} d x\right]$
$=\frac{e^{2 x}}{4(2 x+1)}+C, C$ is integrating constant.

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