A question asked by Vijay

Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}\;dx $

Answer:

$\int \sqrt{\frac{1+x^2}{x^2-x^4}}\;dx $

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

Substitute $x^2 = \cos 2\theta$ and $ 2x\;dx = -2\sin 2\theta \; d\theta $

An we have,

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

$=\int \sqrt{\frac{1+ \cos 2\theta}{1- \cos 2\theta}}\; \left(\frac{-\sin 2\theta}{\cos 2\theta} \right)\; d\theta$

$= -\int \sqrt{\frac{2\cos^2 \theta}{2\sin^2 \theta}}\ ; \tan 2\theta\; d\theta$

$= -\int \cot \theta\; \tan 2\theta\; d\theta$

$= -\int \cot \theta\; \frac{2tan \theta}{1-\tan^2 \theta} \; d\theta$

$= -\int \frac{2}{1-\tan^2 \theta} \; d\theta$

$= -\int \frac{2\cos^2 \theta}{\cos^2 \theta -\sin^2 \theta} \; d\theta$

$= -\int \frac{2\cos^2 \theta}{\cos 2\theta} \; d\theta$

$= -\int 2 \cos^2 \theta \sec 2\theta \; d\theta$

$= -\int (1 + \cos 2\theta) \sec 2\theta \; d\theta$

$= -\int (1+ \sec 2\theta) \; d\theta$

$=\frac{1}{2} \ln(\sec 2\theta - \tan 2\theta) - \theta + C$

$= \frac{1}{2} \left( \ln \left( \frac{1}{x^2} - \frac{\sqrt{1-x^4}}{x^2} \right)  -  \cos^{-1} x^2 \right)+ C $

$= \frac{1}{2} \left( \ln \left(  \frac{1 - \sqrt{1-x^4}}{x^2} \right)  -  \cos^{-1} x^2 \right)+ C $

Credit: Sourav De provided the first correct answer, which is a little different than here.