${\displaystyle \int \sqrt{\frac{e^x-1}{e^x+1}}}dx$ is equal to

${\displaystyle \int \sqrt{\frac{e^x-1}{e^x+1}}dx={\displaystyle \int \sqrt{\frac{\sec \theta -1}{\sec \theta +1}}}.\tan \theta d\theta }$

                $\left[Put{e}^x=\sec \theta \Rightarrow dx=\tan \theta d\theta \right]$

$={\displaystyle \int \sqrt{\frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}}\left(\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{\cos \theta }\right)}d\theta ={\displaystyle \int \frac{\left(1-\cos \theta \right)}{\cos \theta }}d\theta$

$={\displaystyle \int \left(\sec \theta -1\right)d\theta =\log \left(\sec \theta +\tan \theta \right)-\theta +C}$

$=\log \left(e^x+\sqrt{e^{2x}-1}\right)-{\sec }^{-1}\left(e^x\right)+C$