Question
Integrate the function $\frac{x e^{x}}{(1+x)^{2}}$

Answer

$I=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{\frac{x}{(1+x)^{2}}\right\} d x$ 
= $\int e^{x}\left\{\frac{1+x-1}{(1+x)^{2}}\right\} d x$ 
= $\int e^{x}\left\{\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right\} d x$ 
Now,
Let $\frac{1}{1+x}=f(x) \Rightarrow f^{\prime}(x)=-\frac{1}{(1+x)^{2}}$ 
We know that,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C$
Thus,
$\int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+C$

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