Question

$\int \frac{e^{2x}}{1+e^x}dx$

Answer 1

$$\begin{gathered} \int \frac{e^{2x}dx}{1+e^x} \\ \Rightarrow \int \frac{e^x.e^x}{1+e^x}dx \\ \mathrm{Let}1+e^x=t \\ \Rightarrow e^x=t-1 \\ \Rightarrow e^{x}dx=dt \\ \mathrm{Now},\int \frac{e^x.e^x}{1+e^x}dx \\ =\int \frac{\left(t-1\right).dt}{t} \\ =\left(1-\frac{1}{t}\right)dt \\ =t-\log \left|t\right|+C \\ =\left(1+e^x\right)-\log \left(1+e^x\right)+C \\ \mathrm{Let}C+1=C\text{'} \\ =e^x-\log \left(1+e^x\right)+C\text{'} \end{gathered}$$