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

$\frac{e^x}{x+1}+c$ $\frac{e^x}{x-1}+c$ $\frac{x}{x+1}+c$ $\frac{x}{x-1}+c$

$\frac{e^x}{x+1}+c$

$\int \frac{x e^x}{(x+1)^2} d x=\int e^x\left(\frac{(x+1)}{(x+1)^2}-\frac{1}{(x+1)^2}\right) d x$ $=\int e^x [\underbrace{\frac{1}{(x+1)}}_{f(x)}-\underbrace{\frac{1}{(x+1)^2}}_{f'(x)} ] d x$ it of the form f(x) so  $\int e^x (f(x) + f'(x) dx = e^x f(x) + c$ $=\frac{e^x}{x+1}+c$ Option: 1