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Let $I=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{\frac{x}{(1+x)^{2}}\right\} d x$

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

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

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

$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x$

It is known that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+\mathrm{C}$

$\therefore \int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+\mathrm{C}$