Evaluate $$I=\int_0^1 \frac{\ln(1-x^2)\arcsin^2 x}{x^2} {\rm d}x.$$
Maybe, we can make a substitution $x=:\sin u, u\in [0,\pi/2]$. Then $$I=2\int_0^{{\pi}/2}\frac{u^2\cos u\ln\cos u}{\sin^2 u}{\rm d}u.$$ Can we go on from here?
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WA says this here $$\frac{1}{18} \sqrt{\pi } \left(-\frac{18 \Gamma \left(-\frac{1}{4}\right) , _3F_2\left(\frac{1}{4},\frac{3}{4},1;\frac{5}{4},\fr ac{5}{4};1\right)}{\Gamma \left(\frac{1}{4}\right)}-\frac{2 \Gamma \left(\frac{1}{4}\right) \left(4 , _3F_2\left(\frac{3}{4},1,\frac{5}{4};\frac{7}{4},\fr ac{7}{4};1\right)+\log (512)\right)}{\Gamma \left(-\frac{1}{4}\right)}-9 \sqrt{\pi } \log (4)-\frac{36 \sqrt{2} \pi ^2}{\Gamma \left(-\frac{1}{4}\right)^2}+\frac{36 \pi \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(-\frac{1}{4}\right)}\right)$$ – Dr. Sonnhard Graubner May 18 '19 at 15:59
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Where did you find this problem? Why do you think it has a nice solution? – John Doe May 18 '19 at 16:37
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Hint. $$\int_0^1 \frac{\ln(1-x^2)}{x^2}\arctan^2(x)dx=-\sum_{n=0}^\infty \frac{1}{n}\int_0^1 x^{2n}\arctan^2(x)dx.$$ – user659895 May 18 '19 at 15:56