Evaluate $\int \frac{dx}{(1-x^2)\sqrt[4]{(2x^2-1)}}$

Question

How can we evaluate the following integral?

$$ \mathcal{I} = \int \frac{dx}{(1-x^2)\sqrt[4]{(2x^2-1)}} $$

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Some notes...

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

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

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

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

Answer 1

Substitute $t=\sqrt[4]{2x^2-1}$ to simplify the integral first

$$ I= \int \frac{1}{(1-x^2)\sqrt[4]{2x^2-1}}dx = \frac1{\sqrt2}\int \frac{4 t^2}{(1-t^4)\sqrt{1+t^4}}dt $$ and break up the integrand $$ \frac{4 t^2}{(1-t^4)\sqrt{1+t^4}} = \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}+ \frac{ t^2+1}{(1-t^2)\sqrt{1+t^4}} $$ 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 $$\int \frac{t^2-1}{(1+t^2)\sqrt{1+t^4}}dt = \int \frac{d(t+\frac1t)}{(t+\frac1t)\sqrt{(t+\frac1{t})^2-2}} $$

Answer 2

Mathematica gives $$I=\frac 12\left(\tanh ^{-1}\left(\frac{x}{\sqrt[4]{2 x^2-1}}\right)+\tan ^{-1}\left(\frac{x}{\sqrt[4]{2 x^2-1}}\right)\right)+C$$