How to calculate this integral with square roots: $\int\frac{ \sqrt{x+1} }{ \sqrt{ x-1 }} \, dx$

Question

How would you calculate this integral:

$$\int_{}\frac{ \sqrt{x+1} }{ \sqrt{ x-1 }} \, dx$$

Answer 1

HINT:

$$\sqrt{\dfrac{x+1}{x-1}}=u\implies x=\dfrac{u^2+1}{u^2-1}=1+\dfrac2{u^2-1}$$

and use Partial Fraction Decomposition

Answer 2

Hint: Multiply top and bottom by $\sqrt{x+1}$. So we want $$\int \frac{x+1}{\sqrt{x^2-1}}\,dx.$$ Now let $x=\cosh t$.

Answer 3

HINT:

Use $\sqrt{\dfrac{x+1}{x-1}}=\tan y$

$\implies x=-\sec2y\implies dx=-2\sec2y\tan2y\ dy$

and $\tan2y=\dfrac{2\tan y}{1-\tan^2y}=?$

Now, $$\int\sqrt{\dfrac{x+1}{x-1}}dx=-\int2\sec2y\tan2y\tan y\ dy$$

$\sec2y\tan2y\tan y=\dfrac{2\sin^2y}{\cos^22y}=\dfrac{1-\cos2y}{\cos^22y}=\sec^22y-\sec2y$

Hope you can take it from here