If $\int \frac{dx}{1+e^x}=x+f\left(x\right)+C,$then $f\left(x\right)$ is equal to 

Let $1+e^x=t\Rightarrow e^x=t-1\Rightarrow x=\log (t-1)$

$\text{ so }dx=\frac{1}{t-1}dt$

Thus $I=\int \frac{dx}{1+e^x}=\int \frac{dt}{t(t-1)}=\int \left(\frac{1}{t-1}-\frac{1}{t}\right)dt$

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