$I=\int \frac{1}{\sqrt{1-e^{2 x}}} d x$ is equal to

$\ln \left|\tan \left(\frac{\sin ^{-1} e^x}{2}\right)\right|+c$

$e^{x}=t$

$e^{x} dx=dt$

$\int \frac{d t}{t \sqrt{1-t^2}}$, put  $\sin ^{-1} t=z$

$\frac{1}{\sqrt{1-t^2}} d t=d z=\int \frac{d z}{\sin z}=\int ~cosec~z d z=\ln \left|\tan \frac{z}{2}\right|+k$

$=\ln \left|\tan \left(\frac{\sin ^{-1} e^x}{2}\right)\right|+k$

Hence (3) is the correct answer.