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Question

Let $I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0$. If $\lim_\limits{x \rightarrow \infty} I(x)=0$, then $I(1)$ is equal to :

$\frac{e+1}{e+2}-\log _{e}(e+1)$
$\frac{e+1}{e+2}+\log _{e}(e+1)$
$\frac{e+2}{e+1}-\log _{e}(e+1)$
$\frac{e+2}{e+1}+\log _{e}(e+1)$

Solution

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