Question

Show that $\int \sqrt{\left(\frac{x+1}{x-1}\right)}dx=\sqrt{x^2-1}+\log [x+\sqrt{x^2-1}]$.

Answer 1

Let $I=\int \sqrt{\frac{x+1}{x-1}}dx$
Put $t=\frac{x+1}{x-1}\Rightarrow (\frac{1}{x-1}-\frac{x+1}{{(x-1)}^2})dx$
$I=-2\int \frac{\sqrt{t}}{{(1-t)}^2}dt$
Put $u=\sqrt{t}\Rightarrow du=\frac{1}{2\sqrt{t}}dt$
$I=-4\int \frac{u^2}{{(1-u^2)}^2}du$
$=-4\int (-\frac{1}{4(u+1)}+\frac{1}{4{(u+1)}^2}+\frac{1}{4(u-1)}+\frac{1}{4{(u-1)}^2})du$
$=\int \frac{1}{u+1}du-\int \frac{1}{{(u+1)}^2}du-\int \frac{1}{u-1}du-\int \frac{1}{{(u-1)}^2}du$
$=\log (u+1)+\frac{1}{u+1}-\log (u-1)+\frac{1}{u-1}$
$=\log (\sqrt{t}+1)+\frac{1}{\sqrt{t}+1}-\log (\sqrt{t}-1)+\frac{1}{\sqrt{t}-1}$
$=\log (\sqrt{\frac{x+1}{x-1}}+1)+\frac{1}{\sqrt{\frac{x+1}{x-1}}+1}-\log (\sqrt{\frac{x+1}{x-1}}-1)+\frac{1}{\sqrt{\frac{x+1}{x-1}}-1}$
$=\sqrt{x^2-1}+\log (x+\sqrt{x^2-1})$