Answer
\[e^x-\ln|1+e^x|+C\]
Work Step by Step
Let \[I=\int\frac{e^{2x}}{1+e^x}dx\]
\[I=\int\frac{e^{x}}{1+e^x}\cdot e^xdx\]
Substitute $1+e^x=t\;\;\;\ldots (1)$
$\;\;\;\;\;\;\;\;\;\Rightarrow dt=e^x dx$
\[I=\int\frac{(t-1)}{t}dt\]
\[I=\int\left(1-\frac{1}{t}\right)dt\]
\[I=t-\ln|t|+C_1\]
Where $C_1$ is constant of intergration
From (1)
\[I=(1+e^x)-\ln|1+e^x|+C\]
\[I=e^x-\ln|1+e^x|+C\]
Where $C=1+C_1$
Hence \[I=e^x-\ln|1+e^x|+C.\]
