$\text{ Let }\mathrm{I}\left(\mathrm{x}\right)=\int \frac{(\mathrm{x}+1)}{\mathrm{x}{\left(1+{\mathrm{xe}}^{\mathrm{x}}\right)}^2}\mathrm{dx},\mathrm{x}>0\text{. If }\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \mathrm{I}\left(\mathrm{x}\right)=0\text{, then }\mathrm{I}\left(1\right)\text{ is equal to }$

$$\begin{gathered} \begin{array}{l}1+{\mathrm{xe}}^{\mathrm{x}}=\mathrm{t}\Rightarrow {\mathrm{e}}^{\mathrm{x}}(\mathrm{x}+1)\mathrm{dx}=\mathrm{dt}\\ \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dt}}{(\mathrm{t}-1){\mathrm{t}}^2} \\ =\int \left(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2}\right)\mathrm{dt} \\ =\ln (t-1)-\ln t+\frac{1}{t}+C\\ =\ln \left|{\mathrm{xe}}^{\mathrm{x}}\right|-\ln \left|1+{\mathrm{xe}}^{\mathrm{x}}\right|+\frac{1}{1+{\mathrm{xe}}^{\mathrm{x}}}+C\\ =\ln \left|\frac{{\mathrm{xe}}^{\mathrm{x}}}{1+{\mathrm{xe}}^{\mathrm{x}}}\right|+\frac{1}{1+{\mathrm{xe}}^{\mathrm{x}}}+C(=0)\end{array} \end{gathered}$$
$$\begin{gathered} \underset{x\to \infty }{\lim }I\left(x\right)=\underset{x\to \infty }{\lim }\ln \left|\frac{x{e}^x}{1+x{e}^x}\right|+\frac{1}{1+x{e}^x}+C=0 \\ \Rightarrow C=0 \\ I\left(1\right)=\ln \left|\frac{e}{1+e}\right|+\frac{1}{1+e} \\ =1-\ln (1+e)+\frac{1}{1+e} \\ =\frac{e+2}{e+1}-\ln (1+e) \end{gathered}$$