$I=\int \frac{(1+x)}{x\left(1+x e^x\right)^2} d x$ is equal to :

$\ln \left(\frac{x e^x}{1+x e^x}\right)+\left(\frac{1}{1+x e^x}\right)+c$

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

$I=\int \frac{d p}{(p-1) p^2}$

$\frac{1}{(p-1) p^2}=\frac{A}{(p-1)}+\frac{B}{p}+\frac{C}{p^2}$

$1=Ap^2+B(p)(p-1)+C(p-1)$

For $p=1, p=0$, and $p=-1, A=1, C=-1$ and $B=-1$.

$\Rightarrow I=\int \frac{1}{(p-1)} d p-\int \frac{d p}{p}-\int \frac{d p}{p^2}=\ln \frac{(p-1)}{p}+\frac{1}{p}+c=\ln \left(\frac{x e^x}{1+x e^x}\right)+\left(\frac{1}{1+x e^x}\right)+c$

Hence (4) is the correct answer.