Think about $e^x$ ?

$\large \displaystyle \int$ $\dfrac{(x+1-1)e^x}{(1+x)^2}dx$ (adding and subtracting 1)

$\large \displaystyle \int$ $e^x \Bigg(\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\Bigg)dx$

The above integral is of the form, $\int e^x\big(f(x)+f'(x)\big)dx=e^xf(x)$ where $f(x)=\dfrac{1}{x+1}$

$\therefore$ The answer is $\large \dfrac{e^x}{x+1}$

We need to integrate $\large \displaystyle \int \dfrac{xe^x}{(1+x)^2} \, dx = \, ?$

Let $u= x+1, \implies du= dx+1$ ;

our integral will become $\large \int \dfrac{xe^x}{(1+x)^2} \, dx =$ $e^{-1}\large\int\frac{(u-1)e^{u}}{u^{2}}$ $du$ ;

Now solving $\large\int\frac{(u-1)e^{u}}{u^{2}}du$ ;

Rewrite: $\large\int \frac{e^{u}}{u}$ $-\large\frac{e^{u}}{u^{2}}du$ ;

Apply linearity: $\large\int\frac{e^{u}}{u}$ du - $\large\int\frac{e^{u}}{u^{2}}$ du;

For the integral $\large\int\frac{e^{u}}{u^{2}}$ $du$ , we integrate by parts: $\int f'g= fg- \int fg'$ ;

$\large f'= \frac{1}{u^{2}} , g$ = $\large e^{u}$ $\implies$ $f=\frac{-1}{u} , g'$ = $\large e^{u}$ ;

$\large\int\frac{e^{u}}{u}$ $du$ $-\large \int f'g= \large\int\frac{e^{u}}{u}$ $du$ $+\large fg+ \int fg'$ = $\large\int\frac{e^{u}du}{u}+(\large\int\frac{-e^{u}du}{u}+\large\frac{e^{u}}{u})$ ; the signs changed because of the negative multiplication.

The integral $\large\frac{e^{u}du}{u}$ cancels, leaving only $\large\frac{e^{u}}{u}$ ;

Plug in solved integrals $\large e^{-1}\int\frac{(u-1)e^{u}}{u^{2}}$ $du$ = $\large\frac{e^{u-1}}{u}$ ;

Undo substitution $u=x+1$ yields in $\large\frac{e^{x}}{x+1}$ $\implies$ $\large\int\frac{xe^{x}dx}{(x+1)^{2}}$ = $\large\frac{e^{x}}{(x+1)}$ $+C$ .

Highly unorthodox method..but if its a speed-test then look at the fourth option.. By seeing it you can tell that derivative of that( quotient rule) will give you the question...waiting for someone to post the methodical answer...