Question

$\int \sqrt{\left(\frac{x-1}{x+1}\right)}dx$

Answer 1

$\int \left(\sqrt{\frac{x-1}{x+1}}\right)dx$

This can be written as

$\Rightarrow \int \sqrt{\frac{(x-1)(x-1)}{(x+1)(x-1)}}dx$

$\to \int \sqrt{\frac{(x-1{)}^2}{x^2-1}}dx$

$\Rightarrow \int \frac{x-1}{\sqrt{x^2-1}}dx$

$\Rightarrow \int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$

In the first interval take $x^2-1=t\Rightarrow 2xdx=dt$

Second interval is in the form of $\int \frac{1}{\sqrt{x^2-a^2}}dx=\log |\sqrt{x^2-a^2}+x|+C$

Using these results we get

$\Rightarrow \frac{1}{2}\int \frac{dt}{\sqrt{t}}-\log |\sqrt{x^2-1}+x|+C$

$\Rightarrow \left(\sqrt{t}\right)-\log |\sqrt{x^2-1}+x|+C$

$\Rightarrow \left(\sqrt{x^2-1}\right)-\log |\sqrt{x^2-1}+x|+C$