<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] Recall the half-angle identity
$$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \tag{1}$$  This suggests the substitution $$x^2 = \cos 2\theta, \quad x \, dx = -\sin 2\theta \, d\theta$$ which yields for $0 < x < 1$
$$\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}} \, dx
&= \int \frac{1}{x^2} \sqrt{\frac{1+x^2}{1-x^2}} \cdot  x \, dx  \\
&= \int \sec 2\theta \cot \theta \,(-\sin 2\theta) \, d\theta \\
&= - \int \tan 2\theta \cot \theta \, d\theta \\
&= - \int (1 + \sec 2\theta) \, d\theta \\
&= - \theta - \frac{1}{2} \log (\sec 2\theta + \tan 2\theta) + C \\
&= - \frac{1}{2} \arccos x^2 - \frac{1}{2} \log \left( \frac{1 + \sqrt{1-x^4}}{x^2} \right) + C.
\end{align} $$
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