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How do I compute the next integral:

$$\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx \;?$$

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Let $\sqrt{\dfrac{x-1}{x+1}} = t$. We then get that $$\dfrac{x-1}{x+1} = t^2 \implies x-1 = t^2(x+1) \implies x = \dfrac{1+t^2}{1-t^2} \implies dx = \dfrac{4t}{(1-t^2)^2}dt$$ Hence, we get that $$\int \sqrt{\dfrac{x-1}{x+1}} dx = \int \dfrac{4t^2}{(1-t^2)^2} dt$$ I trust you can take it from here via the method of partial fractions.

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  • $\begingroup$ Thank you, couldn't see that trick. $\endgroup$ – StationaryTraveller Apr 10 '13 at 17:03
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As for calculus, $x\ge1$ or $x<-1$

WLOG $x=\sec2u,0\le2u<\pi$

$\tan2u=\pm\sqrt{x^2-1}$

The plus sign needs to be considered if $\tan2u\ge0$ if $0\le2u\le\dfrac\pi2$

$$I=\int\sqrt{\dfrac{x-1}{x+1}}dx=2\int\tan u\sec2u\tan2u\ du$$

$$=\int\dfrac{4\sin^2u}{\cos^22u}du$$

As $\cos2t=1-2\sin^2t,$

$$I=2\int(\sec^22u-\sec2u)du$$

Use How do I integrate $\sec(x)$?

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We cannot have $-1\le x\le1$

Let $\displaystyle I=\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx ,J=\int \sqrt{\frac{x+1}{x-1}}\,\mathrm dx$

$\displaystyle I+J=\int\frac{|x-1|+|x+1|}{\sqrt{x^2-1}}\,\mathrm dx, I-J=\int\frac{|x-1|-|x+1|}{\sqrt{x^2-1}}\,\mathrm dx$

Case $\#1:$ If $\displaystyle x>1,I+J=\int\frac{x-1+x+1}{\sqrt{x^2-1}}\,\mathrm dx, I-J=\int\frac{x-1-x+1}{\sqrt{x^2-1}}\,\mathrm dx$

For $\displaystyle I+J,$ set $\sqrt{x^2-1}=u$

For $\displaystyle I-J,$ set $x=\sec t$

Case $\#2:$ Check if $x<-1?$

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