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

How would you calculate this integral:

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

• Are the $(1/2)$'s supposed to be powers? If so, you should enclose them in braces to get $\LaTeX$ to render them properly. 1^{(1/2)} gives $1^{(1/2)}$, in contrast to 1^(1/2) which gives $1^(1/2)$ This was before the edit using the square root signs, but +1 for MathJax – Ross Millikan Apr 12 '16 at 4:59
• It's fixed now, sorry for the error. – user43438 Apr 12 '16 at 5:00

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

• I wouldn't call that a "hint"; I'd call it solution with the finer details omitted. – Michael Hardy Apr 12 '16 at 5:03
• For a beginner it's a hint. I still don't know what happens with $du$ :) – user43438 Apr 12 '16 at 5:05
• @user43438, Apply derivative to find $$dx=\dfrac{-4u}{(u^2-1)^2}du$$ – lab bhattacharjee Apr 12 '16 at 12:40

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

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