Answer
$$\displaystyle\int{\sqrt{\frac{x+1}{x-1}}dx}=\sqrt{x^2-1}+\ln \left| \sqrt{x^2-1}+x \right|+C$$
Work Step by Step
$I=\displaystyle\int{\sqrt{\frac{x+1}{x-1}}dx}$
$\left[ \begin{matrix}
m=\sqrt{x-1}& \sqrt{m^2+2}=\sqrt{x+1}& dx=2\sqrt{x-1}\ dm\\
\end{matrix} \right] $
$I=\displaystyle\int{\sqrt{\frac{x+1}{x-1}}\times 2\sqrt{x-1}\,\,dm}\\
I=\displaystyle2\int{\sqrt{x+1}\,\,dm}\\
I=\displaystyle2\int{\sqrt{m^2+2}\ dm}$
$\left[ \begin{matrix}
m=\sqrt{2}\tan \theta& m^2=2\tan ^2\theta& dm=\sqrt{2}\sec ^2\theta \ d\theta\\
\end{matrix} \right] $
$I=\displaystyle2\int{\sqrt{2\tan ^2\theta +2}\,\,\times \sqrt{2}\sec ^2\theta \ d\theta}\\
I=\displaystyle4\int{\sqrt{\tan ^2\theta +1}\,\,\times \sec ^2\theta \ d\theta}\\
I=\displaystyle4\int{\sec ^3\theta \ d\theta}\\
I=\displaystyle\int{4\sec \theta \times \sec ^2\theta \,\,d\theta}$
$\left[\begin{matrix}
u=\sec \theta& du=4\sec \theta \tan \theta\\
dv=\sec ^2\theta& v=\tan \theta\\
\end{matrix}\right]$
$\displaystyle4\int{\sec ^3\theta \,\,d\theta}=4\sec \theta\tan \theta-4\int{\sec \theta\tan ^2\theta\,\,d\theta}\\
\displaystyle4\int{\sec ^3\theta \,\,d\theta}=4\sec \theta \tan \theta -4\int{\sec \theta \left( \sec ^2\theta -1 \right) \,\,d\theta}\\
\displaystyle4\int{\sec ^3\theta \,\,d\theta}=4\sec \theta \tan \theta -4\int{\sec ^3\theta -\sec \theta \ d\theta}\\
\displaystyle4\int{\sec ^3\theta \,\,d\theta}=4\sec \theta \tan \theta -4\int{\sec ^3\theta \,\,d\theta +4\int{\sec \theta \,\,d\theta}}\\
\displaystyle8\int{\sec ^3\theta \,\,d\theta}=4\sec \theta \tan \theta +4\int{\sec \theta \,\,d\theta}\\
\displaystyle8\int{\sec ^3\theta \,\,d\theta}=4\sec \theta \tan \theta +4\ln \left| \tan \theta +\sec \theta \right|+C\\
\displaystyle\int{\sec ^3\theta \,\,d\theta}=\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln \left| \tan \theta +\sec \theta \right|+C\\
\displaystyle4\int{\sec ^3\theta \,\,d\theta}=2\sec \theta \tan \theta +2\ln \left| \tan \theta +\sec \theta \right|+C\\
I=\displaystyle2\sec \theta \tan \theta +2\ln \left| \tan \theta +\sec \theta \right|+C$
$\left[ \begin{matrix}
\displaystyle\sec \theta =\sqrt{\frac{m^2+2}{2}}& \displaystyle\tan \theta =\frac{m}{\sqrt{2}}\\
\end{matrix} \right] $
$I=\displaystyle2\sqrt{\frac{m^2+2}{2}}\times \frac{m}{\sqrt{2}}+2\ln \left| \frac{m}{\sqrt{2}}+\sqrt{\frac{m^2+2}{2}} \right|+C\\
I=\displaystyle m\sqrt{m^2+2}+2\ln \left| \frac{m+\sqrt{m^2+2}}{\sqrt{2}} \right|+C\\
I=\displaystyle\sqrt{x-1}\sqrt{x+1}+\ln \left| \left( \frac{\sqrt{x-1}+\sqrt{x+1}}{\sqrt{2}} \right) ^2 \right|+C\\
I=\displaystyle\sqrt{x^2-1}+\ln \left| \frac{2\sqrt{x^2-1}+2x}{2} \right|+C\\
I=\displaystyle\sqrt{x^2-1}+\ln \left| \sqrt{x^2-1}+x \right|+C
$
