∫xex(1+x)2dx is equal to

A
$\frac{e^{x}}{1 + x} + C$
B
$\frac{- 1}{1 + x} + C$
C
$\frac{- e^{x}}{(1 + x)} + C$
D
$\frac{e^{x}}{(1 + x)^{2}} + C$

Solution:

Let $\begin{aligned} I = \int \frac{{xe}^{x}}{(1 + x)^{2}} dx = \int e^{x} \frac{(x + 1) - 1}{(1 + x)^{2}} dx \\ = \int e^{x} [\frac{1}{(1 + x)} - \frac{1}{(1 + x)^{2}}] dx \end{aligned}$

Let $f (x) = \frac{1}{1 + x} \Rightarrow f^{'} (x) = - \frac{1}{(1 + x)^{2}}$

we know that $\int e^{x} \{f (x) + f^{'} (x)\} dx = e^{x} f (x)$

$\Rightarrow I = \int e^{x} \{\frac{1}{1 + x} - \frac{1}{(1 + x)^{2}}\} dx = \frac{e^{x}}{1 + x} + C$

$\int \frac{{xe}^{x}}{(1 + x)^{2}} dx$ is equal to

Solution:

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