The value of \(\int \frac{x+1}{x\left(1+x e^x\right)^2} \mathrm{~d} x\) is equal to

\(\begin{aligned} & \quad \text { Let } \mathrm{I}=\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=\int \frac{\mathrm{e}^x(x+1)}{\mathrm{e}^x \cdot x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x \\ & \\ & \quad \text { Put } x \cdot \mathrm{e}^x=\mathrm{t} \\ & \quad \Rightarrow(x+1) \mathrm{e}^x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \\ & \mathrm{I}=\int \frac{\mathrm{d}}{\mathrm{t}(1+\mathrm{t})^2}\end{aligned}\)
\(\begin{aligned} & =\int \frac{1+\mathrm{t}-\mathrm{t}}{\mathrm{t}(1+\mathrm{t})^2} \mathrm{dt} \\ & =\int \frac{1 \mathrm{dt}}{\mathrm{t}(1+\mathrm{t})}-\int \frac{1}{(1+\mathrm{t})^2} \mathrm{dt} \\ & =\int \frac{1+\mathrm{t}-\mathrm{t}}{\mathrm{t}(1+\mathrm{t})}-\int \frac{1}{(1+\mathrm{t})^2} \mathrm{dt} \\ & =\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{(\mathrm{t}+1)} \mathrm{dt}-\int \frac{1}{(1+\mathrm{t})^2} \mathrm{dt} \\ & =\log \mathrm{t}-\log (1+\mathrm{t})+\frac{1}{(1+\mathrm{t})}+\mathrm{c} \\ & =\log x \mathrm{e}^x-\log \left(1+x \mathrm{e}^x\right)+\frac{1}{\left(1+x \mathrm{e}^x\right)}+\mathrm{c} \\ \mathrm{I} & =\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}\end{aligned}\)