Question

$\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$ equals to
(where $C$ is constant of integration)

Answer 1

$\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx=\int \frac{1+x^2}{x\sqrt{1-x^4}}dx=\int \frac{dx}{x\sqrt{(1-x^4)}}+\int \frac{xdx}{\sqrt{(1-x^4)}}=\int \frac{x^{3}dx}{x^4\sqrt{1-x^4}}+\int \frac{xdx}{\sqrt{(1-x^4)}}=-\frac{1}{2}\int \frac{udu}{(1-u^2)u}+\frac{1}{2}\int \frac{dv}{\sqrt{(1-v^2)}}$
(putting $1-x^4=u^2,-4x^{3}dx=2udu$ in the first integral and $x^2=v,2xdx=dv$ in the second integral)
$=\frac{1}{2}\int \frac{du}{u^2-1}+\frac{1}{2}{\sin }^{-1}v=\frac{1}{2}\times \frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|+\frac{1}{2}{\sin }^{-1}v+C=\frac{1}{4}\ln \left|\frac{\sqrt{(1-x^4)}-1}{\sqrt{(1-x^4)}+1}\right|+\frac{1}{2}{\sin }^{-1}\left(x^2\right)+C$