[Largedisplaystyle int_0^1 dfrac{operatorname{Li}_2left(dfrac{x}{4}
ight)}{4-x}lnleft(dfrac{1+sqrt{1-x}}{1-sqrt{1-x}}
ight)mathrm{d}x=dfrac{pi^4}{1944}
]
(Largemathbf{Proof:})
[egin{align*} {int_0^1 frac{ ext{Li}_2left(dfrac{x}{4}
ight)}{4-x}lnleft(frac{1+sqrt{1-x}}{1-sqrt{1-x}}
ight) mathrm{d}x} &= frac{1}{4}int_0^1left(sum_{n=1}^inftyfrac{H_{n}^{(2)}x^n}{2^{2n}}
ight)lnleft(frac{1+sqrt{1-x}}{1-sqrt{1-x}}
ight) mathrm{d}x \ &= sum_{n=1}^inftyfrac{H_n^{(2)}}{2^{2(n+1)}}int_0^1 x^n lnleft(frac{1+sqrt{1-x}}{1-sqrt{1-x}}
ight) mathrm{d}x\&=sum_{n=1}^inftyfrac{H_n^{(2)}}{(n+1)2^{2(n+1)}}int_0^1frac{x^n}{sqrt{1-x}}mathrm{d}x \ &= frac{1}{2}sum_{n=1}^inftyfrac{H_n^{(2)}}{(n+1)(2n+1)dbinom{2n}{n}}\&=sum_{n=1}^inftyfrac{H_n^{(2)}}{(n+1)^2dbinom{2(n+1)}{n+1}}\&=frac{2}{3}left(sin^{-1}frac{1}{2}
ight)^4 \ &=Largeoxed{color{blue}{dfrac{pi^4}{1994}}}
end{align*}]