$$ \begin{aligned} & \mathrm{I}=\int \frac{x+1}{x\left(1+x e^x\right)^2} d x \\\\ & \text { Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\\\ & \Rightarrow\left(x e^x+e^x\right) d x=d t \\\\ & \Rightarrow e^x(x+1) d x=d t \end{aligned} $$

$$ \therefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2} $$

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

$$ \Rightarrow 1=\mathrm{A} t^2+(\mathrm{B} t+\mathrm{C})(t-1) $$

Comparing coefficients of $t^2, t$ and constant terms, we get

$$ A+B=0, C-B=0,-C=1 $$

On solving above equations, we get

$$ \begin{aligned} & \mathrm{C}=-1,=\mathrm{B}, \mathrm{A}=1 \\\\ & \therefore \mathrm{I}=\int \frac{1}{t-1} d t+\int \frac{-t-1}{t^2} d t \\\\ & =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t-\int \frac{1}{t^2} d t \\\\ & =\log |t-1|-\log |t|+\frac{1}{t}+\mathrm{C} \\\\ & \Rightarrow \mathrm{I}=\log \left|x e^x\right|-\log \left|1+x e^x\right|+\frac{1}{1+x e^x}+c \end{aligned} $$

$$ =\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C $$

Now, $\lim _{x \rightarrow \infty} \mathrm{I}(x)=0$

$$ \begin{aligned} & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+\mathrm{C}\right\}=0 \\\\ & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{e^x}{\frac{1}{x}+e^x}\right|+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathrm{C}\right\} \\\\ & \Rightarrow 0+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\\\ & \therefore \mathrm{I}(x)=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x} \\\\ & \Rightarrow \mathrm{I}(1)=\log \left|\frac{e}{1+e}\right|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \\\\ & =\frac{2+e}{1+e}-\log |1+e| \end{aligned} $$