# Prove: $\int_{0}^{1}\frac{\ln{x}\,\mathrm{d}x}{\sqrt{x(1-x^2)^2}}\stackrel{?}{=}-\frac18\left[\Gamma{\left(\frac13\right)}\right]^3$

I'd like to evaluate the following definite integral:

$$\int_{0}^{1}\frac{\ln{x}\,\mathrm{d}x}{\sqrt{x(1-x^2)^2}}\stackrel{?}{=}-\frac18\left[\Gamma{\left(\frac13\right)}\right]^3.$$

Preferably, I'd like to see this integral solved without the language of hypergeometric functions. This is integral is very similar to the one posed in this question, and the one in this question. I believe the key to those two integrals is finding some appropriate cubic transformations that turns the integrals into a product of beta functions, but I kept running into problems. My hope is that if the simpler integral I posed above can be solved, the method of solution might be generalizable (in part or whole) towards the solutions of a broader class of integrals of cube-roots of polynomials.

We have: $$I = \frac{1}{4}\int_{0}^{1}z^{-2/3}(1-z)^{-2/3}\log z \,dz$$ hence: $$I =\frac{1}{4}\left. \frac{d}{d\alpha}\left(\int_{0}^{1}z^{-2/3+\alpha}(1-z)^{-2/3}\,dz\right) \right|_{\alpha=0}=\frac{\Gamma(1/3)}{4}\left.\frac{d}{d\alpha}\left(\frac{\Gamma(1/3+\alpha)}{\Gamma(2/3+\alpha)}\right)\right|_{\alpha=0}$$ and by using the identity $\Gamma' = \Gamma\cdot\frac{d}{dz}(\log \Gamma)=\Gamma\cdot\psi$ we have: $$\left.\frac{d}{d\alpha}\left(\frac{\Gamma(1/3+\alpha)}{\Gamma(2/3+\alpha)}\right)\right|_{\alpha=0}=\frac{\Gamma(1/3)}{\Gamma(2/3)}\left(\psi(1/3)-\psi(2/3)\right)$$ so: $$\color{blue}{I = \frac{\Gamma(1/3)^3}{8}\cdot\frac{\sqrt{3}}{\pi}(\psi(1/3)-\psi(2/3))}$$ and the claim follows from the reflection formula for the digamma function.
I have not been able to get rid of hypergeometric functions and I apologize for that. What a CAS found is that $$\int\frac{\ln{x}\,\mathrm{d}x}{\sqrt{x(1-x^2)^2}}=\frac{\left(x \left(x^2-1\right)^2\right)^{2/3}}{\left(1-x^2\right)^{4/3}}A(x)$$ where $$A(x)=\frac{3}{2} \, _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{4}{3};x^2\right) \log (x)-\frac{9}{4} \, _3F_2\left(\frac{1}{3},\frac{1}{3},\frac{2}{3};\frac{4}{3},\frac{4}{3};x^2\right)$$ So, $$\int_{0}^{1}\frac{\ln{x}\,\mathrm{d}x}{\sqrt{x(1-x^2)^2}}=-\frac{\pi ^{3/2} \Gamma \left(\frac{7}{6}\right)}{2^{2/3} \Gamma \left(\frac{2}{3}\right)}=-\frac18\left[\Gamma{\left(\frac13\right)}\right]^3$$