Evaluate $\int \frac{dx}{(1-x^2)\sqrt[4]{(2x^2-1)}}$ How can we evaluate the following integral?
$$
\mathcal{I} = \int \frac{dx}{(1-x^2)\sqrt[4]{(2x^2-1)}}
$$

Some notes...

*

*Any standard integration methods (e.g., substitution method, integration by parts, etc.) does not work.


*I tried using integral-calculator.com, it does not work either.


*Trivially, I tried to set some possible limits (e.g., 0 $\rightarrow$ 1, $-\pi \rightarrow \pi$, etc.) — with no success.


*This is not a homework question. I'm open to the ideas given by experts here.

Substitute $t=\sqrt[4]{2x^2-1}$ to simplify the integral first
$$
I= \int \frac{1}{(1-x^2)\sqrt[4]{2x^2-1}}dx
= 2\sqrt2 \int \frac{t^2}{(1-t^4)\sqrt{1+t^4}}dt
$$
and note that
$$
\frac{4 t^2}{(1-t^4)\sqrt{1+t^4}}
= \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}+ \frac{ t^2+1}{(1-t^2)\sqrt{1+t^4}}
$$
Then, the two resulting integrals are relatively easy to evaluate
\begin{align}
I= &\frac1{\sqrt2}\int \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}+ \frac{ t^2+1}{(1-t^2)\sqrt{1+t^4}}\ dt\\
=&\ \frac12\tan^{-1}\sqrt{\frac{1+t^4}{2t^2}}+ \frac12\coth^{-1}\sqrt{\frac{1+t^4}{2t^2}}+C
\end{align}
For example
$$\int \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}dt =
\int \frac{d(t+\frac1t)}{(t+\frac1t)\sqrt{(t+\frac1{t})^2-2}}
$$
