How to integrate this integral $\int _{0}^{1} \frac{dx}{\sqrt{x(x^2-1)}}$??

Question

I want to integrate this integral $\int _{0}^{1} \frac{dx}{\sqrt{x(x^2-1)}}$.

I found that

\begin{align*} \int _{0}^{1} \frac{dx}{\sqrt{x(x^2-1)}}&=-i\int _{0}^{1} \frac{dx}{\sqrt{x(1-x^2)}}\\ &=-\frac{i}{2}\int _{0}^{1} (1-u)^{-1/2}u^{-3/4}du \end{align*}

I'm getting stuck... I don't know how to integrate this function..

Any help is appreciated!!

Thank you!

Answer 1

$$\int (1-u)^{-1/2}u^{-3/4}\,du=B_u\left(\frac{1}{4},\frac{1}{2}\right)$$ where appears the incomplete beta function.

Using the relations between beta and gamma functions, $$\int _{0}^{1} (1-u)^{-1/2}u^{-3/4}\,du=\sqrt{\pi }\,\,\frac{ \Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}$$