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How can we integrate $$ \int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} \, dx $$

I thought about substituting $u=\frac 1x$ to transform the integral into a slightly-simpler looking one: $$ -\int \sqrt{\frac{1-u}{1+u}} \, du $$ Am I on the right track?

Following @Shobhit's comment, I let $u=\cos (2w)=\cos^2 w - \sin^2 w$, so that $du=-2 \sin(2w) \, dw = -4 \sin w \cos w \, dw$ and so \begin{align} \int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} dx &= -\int \sqrt{\frac{1-u}{1+u}} \, du \\ &= 4 \int |\tan w| \sin w \cos w \, dw \\ &= 4 \int \sin^2 w \, dw \end{align} If this is correct, then I can finish the rest of this integral on my own.

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    Yes. Now let u=cos2x – 2017-01-30
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    @Shobhit Should I let $u=cos 2w$ or some variable other than $x$? Because I already have $x$ to begin with in the problem. – 2017-01-30
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    Yes, ofcourse. As u see fit. – 2017-01-30
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    $$\cos2w=1-2\sin^2w$$ – 2017-01-30

2 Answers 2

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The faster way to integrate such a function is to let $$ u=\sqrt{\frac{x-1}{x+1}} $$ so that $$ x= \frac{u^2+1}{1-u^2} \implies dx=\frac{4 u}{\left(u^2-1\right)^2}du $$ the integral thus becomes $$ \int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} dx=\int u \frac{(1-u^2)^2}{(1+u^2)^2} \frac{4 u}{\left(u^2-1\right)^2}du=\int \frac{4u^2}{(1+u^2)^2}du $$ and now you can use Partial Fraction Decomposition, which goes off easy, you must obtain: $$ \int \frac{4u^2}{(1+u^2)^2}du=2 \arctan(u)-\frac{2 u}{u^2+1}+c=2 \arctan \left(\sqrt{\frac{x-1}{x+1}}\right)-\frac{\sqrt{x^2-1}}{x}+c $$

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After your substitution from $\displaystyle x= \frac{1}{u}$ $$I = -\int\sqrt{\frac{1-u}{1+u}}du = \int\frac{u-1}{\sqrt{1-u^2}}du = \int\frac{u}{\sqrt{1-u^2}}du-\int\frac{1}{\sqrt{1-u^2}}du$$

$$I = -\sqrt{1-u^2}-\arcsin(u)+\mathcal{C} = -\frac{\sqrt{x^2-1}}{x}-\arcsin\left(\frac{1}{x}\right)+\mathcal{C}$$