$I=\int_0^1 \frac{x e^x}{(x+1)^2} d x$
$
\begin{aligned}
I & =\int_0^1 e^x\left[\frac{x}{(x+1)^2}\right] d x \\
& =\int_0^1 e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x
\end{aligned}
$
Let $f(x)=\frac{1}{x+1} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+1)^2}$
$\begin{aligned} \therefore \quad I & =\int_0^1 e^x\left[f(x)+f^{\prime}(x)\right] d x \\ \Rightarrow \quad I & =\left.e^x f(x)\right|_0 ^1 \\ & I=\left[\frac{e^x}{x+1}\right]_0^1=\frac{e}{2}-1\end{aligned}$