\(\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=\)

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