I've conjectured the following closed-form:
I=1∫0((x2+1)arcsin(x)√1−x2+2xln(x2+1))lnxx3+xdx=−2Gln2,
where G is Catalan's constant.
Numerically
I≈−1.2697979381877088371491554851603117320986537271546606092465…
How to prove it?
Answer
Apply the obvious substitution x↦sinx to the first integral
I=π2∫0xln(sinx)sinx dx+1∫02lnxln(1+x2)1+x2 dx
The latter integral has been addressed here and is equivalent to
1∫02lnxln(1+x2)1+x2 dx=4ℑLi3(1−i)−2Gln2+3π316+π4ln22
As for the first integral,
π2∫0xln(sinx)sinx dx=21∫0arctanxln(2x1+x2)x dx=2Ti2(1)ln2+21∫0arctanxlnxx dx−21∫0arctanxln(1+x2)x dx=2Gln2−2∞∑n=0(−1)n(2n+1)3+21∫0lnxln(1+x2)1+x2 dx+41∫0xarctanxlnx1+x2 dx
and the integral
41∫0xarctanxlnx1+x2 dx=−8ℑLi3(1−i)−5π316−π2ln22
has also been established in the link above. (Both integrals were covered in the evaluation of J2.) Hence the closed form is indeed
I=−2Gln2
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