Question

$\int \frac{e^x-1}{e^x+1}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{e^x-1}{e^x+1}dx \\ =\int \frac{2e^x-\left(e^x+1\right)}{e^x+1}dx \\ =\int \frac{2e^x}{e^x+1}dx-\int dx \\ \mathrm{Putting}{e}^x+1=t \\ \Rightarrow e^{x}dx=dt \\ \therefore I=\int \frac{2}{t}dt-\int dx \\ =2\log \left|t\right|-x+C \\ =2\log \left|e^x+1\right|-x+C \\ =2\log \left(e^x+1\right)-x+C \end{gathered}$$