Question

Write a value of $\int \frac{1}{1+e^x}dx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{1+e^x} \\ \mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}{e}^x \\ \Rightarrow I=\int \frac{{\displaystyle \frac{1}{e^x}}dx}{{\displaystyle \frac{1}{e^x}}+1} \\ =\int \frac{e^{-x}dx}{e^{-x}+1} \\ \mathrm{Let}{e}^{-x}+1=t \\ -e^{-x}dx=dt \\ \Rightarrow e^{-x}dx=-dt \\ \therefore I=\int -\frac{dt}{t} \\ =-\log \left|t\right|+C \\ =-\log \left|1+e^x\right|+C\left(\because t=1+{\mathrm{e}}^{\mathit{x}}\right) \end{gathered}$$