Answer
$$\int \frac{d x}{1+e^{x}}=x-\ln \left(e^{x}+1\right)+C $$
Work Step by Step
Given $$\int \frac{d x}{1+e^{x}}$$ \begin{array}{l}{\text { Let } u=e^{x}, \ \ d u=e^{x} d x \ \text { and } d x=\frac{d u}{u} \text { . Then } \\ \begin{aligned}I&=\int \frac{d x}{1+e^{x}}\\ &=\int \frac{d u}{(1+u) u}\\ \text{Since} \ \ \ \frac{1}{u(u+1)} &=\frac{A}{u}+\frac{B}{u+1}\end{aligned} \\ \Rightarrow} \\ {1=A(u+1)+B u \\ \text { Setting } u=-1 \Rightarrow B=-1 . \text { Setting } u=0\Rightarrow A=1. \text { Thus, }} \\ {\begin{aligned}I&=\int \frac{d u}{u(u+1)}\\&=\int\left(\frac{1}{u}-\frac{1}{u+1}\right) d u\\&=\ln |u|-\ln |u+1|+C\\ &=\ln e^{x}-\ln \left(e^{x}+1\right)+C\\ &=x-\ln \left(e^{x}+1\right)+C \end{aligned}}\end{array}