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

Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.

In order to evaluate the integral over all domain $ x\in (-1,0)\cup(0,1)$, substitute $t=x\sqrt{x^2}$. Then, $dt =2\sqrt{x^2}\ dx$ and
\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}}\ dx
 =& \int \frac{1+x^2}{\sqrt{x^2(1-x^4)}}\ dx
=\frac12\int \frac{1+\sqrt{t^2}}{\sqrt{t^2(1-t^2)}}\ dt \\
=& \ \frac12\int \frac{1}{\sqrt{1-t^2}}+ \frac{1}{\sqrt{t^2(1-t^2)}}\ dt\\
=& \ \frac12 \sin^{-1}t-\frac12\sinh^{-1}
\frac{\sqrt{1-t^2}}t \\
= & \ \frac12 \sin^{-1}\left( x\sqrt{x^2}\right)-\frac12\sinh^{-1}\bigg(
\frac1x \sqrt{\frac{1-x^4}{x^2}}\bigg)\\ 
\end{align}
