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How would you calculate this integral:

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

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    $\begingroup$ 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 $\endgroup$ – Ross Millikan Apr 12 '16 at 4:59
  • $\begingroup$ It's fixed now, sorry for the error. $\endgroup$ – user43438 Apr 12 '16 at 5:00
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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

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    $\begingroup$ I wouldn't call that a "hint"; I'd call it solution with the finer details omitted. $\endgroup$ – Michael Hardy Apr 12 '16 at 5:03
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    $\begingroup$ For a beginner it's a hint. I still don't know what happens with $du$ :) $\endgroup$ – user43438 Apr 12 '16 at 5:05
  • $\begingroup$ @user43438, Apply derivative to find $$dx=\dfrac{-4u}{(u^2-1)^2}du$$ $\endgroup$ – lab bhattacharjee Apr 12 '16 at 12:40
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Hint: Multiply top and bottom by $\sqrt{x+1}$. So we want $$\int \frac{x+1}{\sqrt{x^2-1}}\,dx.$$ Now let $x=\cosh t$.

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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

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