$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} d x=$
$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} d x=$
$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|-\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$, where $\mathrm{c}$ is $\mathrm{a}$ constant of integration.
$\log \left|1+x \mathrm{e}^x\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=\int \frac{\mathrm{e}^x(x+1)}{\mathrm{e}^x \cdot x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x \\ & \text { Let } x \cdot \mathrm{e}^x=\mathrm{t} \\ & \Rightarrow(x+1) \mathrm{e}^x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}(1+\mathrm{t})^2} \\ & =\int \frac{1+\mathrm{t}-\mathrm{t}}{\mathrm{t}(1+\mathrm{t})^2} \mathrm{dt} \\ & =\int \frac{1 d t}{t(1+t)}-\int \frac{1}{(1+t)^2} d t \\ & =\int \frac{1+t-t}{t(1+t)}-\int \frac{1}{(1+t)^2} d t \\ & =\int \frac{1}{t} d t-\int \frac{1}{(t+1)} d t-\int \frac{1}{(1+t)^2} d t \\ & =\log \mathrm{t}-\int \frac{\mathrm{dt}}{(\mathrm{t}+1)}-\int \frac{1}{(1+\mathrm{t})^2} \mathrm{dt}+\mathrm{c} \\ & \text { Let } y=1+\mathrm{t} \\ & \Rightarrow \mathrm{d} y=\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\log \mathrm{t}-\int \frac{1}{y} \mathrm{~d} y-\int \frac{1}{y^2} \mathrm{~d} y+\mathrm{c} \\ & =\log \mathrm{t}-\log y+\frac{1}{y}+\mathrm{c} \\ & =\log \mathrm{t}-\log (1+\mathrm{t})+\frac{1}{(1+\mathrm{t})}+\mathrm{c} \\ & =\log x \mathrm{e}^x-\log \left(1+x \mathrm{e}^x\right)+\frac{1}{\left(1+x \mathrm{e}^x\right)}+\mathrm{c} \\ & \therefore \quad \mathrm{I}=\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c} \\ & \end{aligned}$