# How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$

I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$ But I've not been successful in doing so, what I tried is $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx=\int _0^1\left(x\right)'\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$ $$\overset{\operatorname{IBP}}=\ln \left(\zeta (2)\right)+\int _0^1\frac{\ln \left(1-x\right)}{\operatorname{Li}_2\left(x\right)}\:dx$$ I also tried using identities for the dilogarithm but they dont do much.

I'm not sure what to do now, any help will be very well regarded, thank you.

Being myself quite lazy, I should compose Taylor series starting from $$\text{Li}_2(x)=\sum_{n=1}^\infty \frac {x^n}{n^2}$$ and being very patient (this is cheating ! I used a CAS); write $$\log \left(\operatorname{Li}_2\left(x\right)\right)=\log(x)+\sum_{n=1}^\infty a_n x^n$$ where, unfortunately, all $$a_n$$'s are positive. The coefficients make the sequence $$\left\{\frac{1}{4},\frac{23}{288},\frac{23}{576},\frac{50119}{2073600},\frac{33769 }{2073600},\frac{257986487}{21946982400},\frac{130265243}{14631321600},\cdots\right\}$$ which decrease quite fast; halas, nothing being recognized by $$OEIS$$.
This gives $$\int_0^1\log \left(\operatorname{Li}_2\left(x\right)\right)\,dx=-1+\sum_{n=1}^\infty \frac{a_n}{n+1}$$ Unfortunately, the convergence is not very fast $$\left( \begin{array}{cc} p & 1+\sum_{n=1}^p \frac{a_n}{n+1} \\ 10 & -0.826294 \\ 20 & -0.824783 \\ 30 & -0.824456 \\ 40 & -0.824334 \\ 50 & -0.824276 \\ 60 & -0.824243 \\ 70 & -0.824223 \\ 80 & -0.824210 \\ 90 & -0.824201 \\ 100 & -0.824194 \\ \dots & \dots \\ \infty &-0.824166 \end{array} \right)$$