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

$$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,.$$
Therefore, since $\log(1)=0$, we have:
$$\operatorname{Li}_3(1) = \sum_{k=1}^\infty {1 \over k^3} = \zeta(3)$$
$$\operatorname{Li}_3(1)-\log(1)\operatorname{Li}_2(1) = \zeta(3)$$
It remains to show that $$\lim_{x\to0} \operatorname{Li}_3(x)-\log(x)\operatorname{Li}_2(x)=0$$
Note that if $z<1$,
$$\operatorname{Li}_2(z) = \sum_{k=1}^\infty {z^k \over k^2} = z + {z^2 \over 2^2} + {z^3 \over 3^2} + \cdots \\< z + {z \over 2^2} + {z \over 2^2}+ {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 8^2}+\cdots \leq z\sum_{k=0}^\infty {1 \over 2^k} = 2z$$
Since $\log(z)<z$, for $z>1$, we also have $\log(z^2)<2z$ thus $\log(x)<2\sqrt{x}$. Thus $\log(\frac1x)>-2\sqrt{x}$, thus $\log(u)>-2\sqrt{\frac1u}$, thus  $0>\log(u)\operatorname{Li}_2(u)>-2\sqrt{u}$.
We may use the Squeeze Theorem to finish the result.
