Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ and do a comparision with $\zeta(3)$? I know from Wolfram Alpha that $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=1.20206$$ and in the other hand, too from this online tool that
$$\int\frac{\log(x)\log(1-x)}{x}dx=\mathrm{Li}_3(x)-\mathrm{Li}_2(x)\log(x)+constant.$$

Question. I would like made a comparision, and need obtain $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx,$$
  more precisely than $1.20206$. I believe that could be $\zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.
Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ to discard that this value is $\zeta(3)$, Apéry constant, or claim that the equality $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=\zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?

This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.
I've searched in this site about this integral $\int\frac{\log(x)\log(1-x)}{x}dx$, and in Wikipedia about a possible identity between $\zeta(3)$ and particular values of logarithmic integrals $\mathrm{Li}_2(x)$ and $\mathrm{Li}_3(x)$. 
References:
[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant

Let $x = e^{-y}$, we have
$$\int_0^1 \frac{\log x\log(1-x)}{x} dx
= \int_0^1 \frac{(-\log x)}{x} \sum_{n=1}^\infty \frac{x^n}{n} dx
= \sum_{n=1}^\infty \frac{1}{n}\int_0^1 (-\log x) x^{n-1} dx\\
= \sum_{n=1}^\infty \frac{1}{n}\int_0^\infty y e^{-ny} dy
= \sum_{n=1}^\infty \frac{1}{n^3}
= \zeta(3)
$$
Please note that we can switch the order of summation and integration because all the individual terms are non-negative.
