# Computing the integral $\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx$

How do I compute the next integral:

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

## 3 Answers

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.

• Thank you, couldn't see that trick. – StationaryTraveller Apr 10 '13 at 17:03

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

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

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 $$x<-1?$$