$\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$

Question

$\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$

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I tried putting $x=\sin \theta$ and changing the limits from $0$ to $\frac{\pi}{2}$and i got $\int_0^\frac{\pi}{2}\sin^2\theta\ln \sin \theta d\theta$ i cannot further solve it

Answer 1

Hint: Use beta function and show that $$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac{\sqrt{\pi}}{{2}}\dfrac{\Gamma(\frac{\alpha+1}{2})}{\Gamma(\frac{\alpha+2}{2})}$$ then your integral is $$\dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx$$ in $\alpha=2$.