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I want to evaluate the integration $\int \limits x \sqrt{\frac{x-1}{x+1}}dx$ ?

I tried putting $t = \sqrt{\frac{x-1}{x+1}}$ but that did not help !

Is there a quick and smart way to evaluate.

gt6989b
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Ahmad
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3 Answers3

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For the square root to be defined you need for instance $x\ge 1$, this suggest a substitution $x=\cosh(u)$.

The integral becomes $$I=\int\cosh(u)\sinh(u)\sqrt{\frac{\cosh(u)-1}{\cosh(u)+1}}\mathop{du}$$

Now using $\cosh(u)^2-\sinh(u)^2=1$ we get


$\displaystyle\int\frac{\cosh(u)\sinh(u)\sqrt{\cosh(u)^2-1}}{\cosh(u)+1}\mathop{du}=\int\frac{\cosh(u)\sinh(u)^2}{\cosh(u)+1}\mathop{du}=\int\cosh(u)^2-\cosh(u)\mathop{du}$


Just linearise the $\cosh(u)^2$ to get the result $$I=\frac u2-\sinh(u)+\frac 14\sinh(2u)$$


Finally use $\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}\quad$ and $\quad\sinh(2\cosh^{-1}(x))=2x\sqrt{x^2-1}$

To get $$I=\frac 12\cosh^{-1}(x)+\sqrt{x^2-1}\left(\frac x2-1\right)+C$$

zwim
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Note that $$\int x \sqrt{\frac{x-1}{x+1}}\,dx=\int\frac x{x+1}\sqrt{x^2-1}\,dx=\int\left(1-\frac1{x+1}\right)\sqrt{x^2-1}\,dx$$ evaluates to $$\int\sqrt{x^2-1}\,dx-\int\frac{\sqrt{x^2-1}}{x+1}\,dx.$$ For the first integral, use integration by parts from the $f=f\cdot1$ trick and for the second integral, note that $$\int\frac{\sqrt{x^2-1}}{x+1}\,dx=\int\frac{x-1}{\sqrt{x^2-1}}\,dx.$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Inside the radical, multiply top and bottom by $x-1$ to get

$$\int \frac{x^2-x}{\sqrt{x^2-1}} \; dx$$

This wants the usual trig sub $x = \sec \theta$ which yields

$$\int \sec^3\theta - \sec^2 \theta \; d\theta.$$

The $\sec^2 \theta$ is easy. The $\sec^3\theta$ is less easy, but is a standard integral.

B. Goddard
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