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

Write the integrand as $\frac{1}{x}+1-\frac{(x+1)e^x}{1+x{e}^x}$

So the given integral is equal to

$\begin{array}{l}\log x+x-\log \left(1+x{e}^x\right)+C\\ =\log x+\log e^x-\log \left(1+x{e}^x\right)+C\\ =\log \frac{x{e}^x}{1+x{e}^x}+C.\end{array}$