If ${\displaystyle \int \frac{(x+1)dx}{x{(1+x{e}^x)}^2}}=\log f\left(x\right)+\frac{1}{1+x{e}^x}+c$ , then $f\left(x\right)=$

$I={\displaystyle \int \frac{(x+1)dx}{x{(1+x{e}^x)}^2}}$

Putting $(1+x{e}^x)=t\Rightarrow (x+1)e^{x}dx=dt$

$I={\displaystyle \int \frac{dt}{(t-1)t^2}}={\displaystyle \int \left(–\frac{1}{t^2}–\frac{1}{t}+\frac{1}{t–1}\right)}dt$

   $=\frac{1}{t}–\log t+\log (t–1)+c$

   $=\frac{1}{1+x{e}^x}+\log \left(\frac{x{e}^x}{1+x{e}^x}\right)+c$

$\Rightarrow f\left(x\right)=\frac{x{e}^x}{1+x{e}^x}$ .