The integral $I=∫\frac{e^x}{1-e^{2x}}dx$ is equal to :

$\frac{1}{2}\left[-log(1-e^x)+log (1+e^x)\right]+C$

The correct answer is Option (1) → $\frac{1}{2}\left[-log(1-e^x)+log (1+e^x)\right]+C$

$I=∫\frac{e^x}{1-e^{2x}}dx$

let $y=e^x$

$dy=e^xdx$

$I=\int\frac{dy}{1-y^2}⇒\frac{1}{2}\int\frac{(1+y)+(1-y)}{1-y^2}$

$=\frac{1}{2}\int\frac{1}{1-y}+\frac{1}{1+y}dy=\frac{1}{2}(-\log|1-y|+\log|1+y|)+C$

$=\frac{1}{2}(-\log|1-e^x|+\log|1+e^x|)+C$