<s>[INST] 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.
 [/INST] Given integral $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Let's factor out $x^2 $ from the denominator.
$\Rightarrow \int \frac{1}{x}\sqrt{\frac{1+x^2}{1-x^2}}dx$
Now multiply and divide the numerator and the denominator by $\sqrt{1+x^2}$
$\Rightarrow \int \frac{1+x^2}{x\sqrt{1-x^4}}dx$
Now we can split the integral into 2 sub-integrals
$\Rightarrow \int \frac{1}{x\sqrt{1-x^4}}dx+\int \frac{x^2}{x\sqrt{1-x^4}}dx$
$\Rightarrow \int \frac{x^3}{x^4\sqrt{1-x^4}}dx+\int \frac{x}{\sqrt{1-x^4}}dx$
For the first integral, use $1-x^4=u^2$ and for the second, use $x^2=v$. The integrals reduce to
$\Rightarrow \frac{-1}{2}\int \frac{du}{1-u^2}+\frac{1}{2}\int \frac{dv}{\sqrt{1-v^2}}$
$\Rightarrow \frac{1}{2}\int \frac{du}{u^2-1}+\frac{1}{2} \sin ^{-1}v$
$\Rightarrow \frac{1}{4}\log |\frac{u-1}{u+1}| +\frac{1}{2}\sin ^{-1}v+c$
$\Rightarrow \frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
Hence, $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx=\frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
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