A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$ A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.

Prove:
  $$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$


If we let $x\mapsto e^{-x}$ and then use the series of $\log(1-e^{-x})$, we obtain that
$$\underbrace{\int_0^{\infty} \sum_{n=1}^{\infty}\log(4 \pi^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle \zeta(2) \log(4 \pi^2)} -\underbrace{\sum_{n=1}^{\infty}\int_0^{\infty} \log(4 \pi^2+y^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle -2\sum_{n=1}^{\infty}\frac{\operatorname{Ci}(2\pi n)}{n^2}+2\log(2)\zeta(2)+2\log(\pi)\zeta(2)}=\pi^{2} ( 4 \log A - 1 )$$
where in the last integral I used the exponential integral and $1.22a$ from this paper.
Q.E.D.
