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
 A: \begin{align}J&=\int_0^1 \frac{\ln(1-x)\ln x}{x} dx\\
&=-\int_0^1 \left(\sum_{n=1}^\infty \frac{x^{n-1}}{n}\right)\ln x\,dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-1}\ln x\,dx\\
&=\sum_{n=1}^\infty \frac{1}{n^3}\\
&=\zeta(3)
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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You can add this "weird" answer to the excellent
$\texttt{@mrtaurho}$ long list:
\begin{align}
&\bbox[10px,#ffd]{%
\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x} =
\left.{\partial^{2} \over \partial\mu\,\partial\nu}\int_{0}^{1}
{\bracks{\pars{1 - x}^{\mu} - 1}x^{\nu} \over x}\,\dd x\,\right\vert_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}}
\\[5mm] = &\
{\partial^{2} \over \partial\mu\,\partial\nu}\bracks{%
\int_{0}^{1}x^{\nu - 1}\pars{1 - x}^{\mu}\,\dd x -
\int_{0}^{1}x^{\nu - 1}\,\dd x}
_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}}
\\[5mm] = &\
{\partial^{2} \over \partial\mu\,\partial\nu}\bracks{%
{\Gamma\pars{\nu}\Gamma\pars{\mu + 1} \over
\Gamma\pars{\nu + \mu + 1}} - {1 \over \nu}}
_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}}
\\[5mm] = &\
{\partial^{2} \over \partial\mu\,\partial\nu}\braces{{1 \over \nu}\bracks{%
{\Gamma\pars{\nu + 1}\Gamma\pars{\mu + 1} \over
\Gamma\pars{\nu + \mu + 1}} - 1}}
_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}}
\\[5mm] = &\
{1 \over 2}\,\partiald[2]{}{\nu}\braces{\Gamma\pars{\nu + 1}
\partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1} \over
\Gamma\pars{\nu + \mu + 1}}}
_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}}
\\[5mm] = &\
{1 \over 2}\,\partiald[2]{}{\nu}\braces{\Gamma\pars{\nu + 1}
\bracks{-\,{\gamma + \Psi\pars{\nu + 1} \over
\Gamma\pars{\nu + 1}} }}_{\ \nu\ =\ 0^{+}}
\\[5mm] = &\
-\,{1 \over 2}\,\Psi\,''\pars{1} = \bbx{\zeta\pars{3}}
\end{align}
A: An informal argument:  $$\log (1-x) = - \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}, \quad |x| < 1.$$  Then $$\frac{\log x \log(1-x)}{x} = -\sum_{k=0}^\infty \frac{x^k \log x}{k+1},$$ and integrating term by term gives $$\int_{x=0}^1 x^k \log x \, dx = \left[ \frac{x^{k+1} \log x}{k+1} \right]_{x=0}^1 - \int_{x=0}^1 \frac{x^k}{k+1} \, dx = - \frac{1}{(k+1)^2}.$$  Therefore, $$\int_{x=0}^1 \frac{\log x \log(1-x)}{x} \, dx = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3).$$
A: 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.
A: There is a variety of possibilities how to show that this integral indeed equals $\zeta(3)$, i.e. Apéry's Constant. I would like to show some of them
I: Taylor Series Expansion of $\log(1-x)$
As it was first suggested within the comments (and done by FDP) we may expand the $\log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain
\begin{align*}
\int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=\int_0^1\frac{\log(x)}x\left[-\sum_{n=1}^\infty\frac{x^n}n\right]\mathrm dx\\
&=-\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\log(x)\mathrm dx\\
&=-\sum_{n=1}^\infty\frac1n\left[-\frac1{n^2}\right]\\
&=\sum_{n=1}^\infty\frac1{n^3}\\
&=\zeta(3)
\end{align*}
This might be the most straightforward approach possible.
II: Integration By Parts
Choosing $u=\log(1-x)$ and $\mathrm dv=\frac{\log(x)}x$ we can apply Integration By Parts which gives
\begin{align*}
\int_0^1\frac{\log(1-x)\log(x)}x&=\underbrace{\left[\log(1-x)\frac{\log^2(x)}2\right]_0^1}_{\to0}+\frac12\int_0^1\frac{\log^2(x)}{1-x}\mathrm dx\\
&=\frac12\int_0^1\log^2(x)\left[\sum_{n=0}^\infty x^n\right]\mathrm dx\\
&=\frac12\sum_{n=0}^\infty\int_0^1x^n\log^2(x)\mathrm dx\\
&=\frac12\sum_{n=0}^\infty\left[\frac2{(n+1)^3}\right]\\
&=\sum_{n=1}^\infty\frac1{n^3}\\
&=\zeta(3)
\end{align*}
Again, we utilized a series expansion, this time the one of the geometric series.
III: Integral Representation of the Zeta Function
To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $\log(x)\mapsto -x$ followed by Integration By Parts again we find
\begin{align*}
\int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=-\int_\infty^0(-x)\log(1-e^{-x})\mathrm dx\\
&=-\int_0^\infty x\log(1-e^{-x})\mathrm dx\\
&=\underbrace{\left[\frac{x^2}2\log(1-e^{-x})\right]_0^\infty}_{\to0}+\frac12\int_0^\infty\frac{x^2}{1-e^{-x}}e^{-x}\mathrm dx\\
&=\frac1{\Gamma(3)}\int_0^\infty\frac{x^{3-1}}{e^x-1}\mathrm dx\\
&=\zeta(3)
\end{align*} 
Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.
IV: The Trilogarithm $\operatorname{Li}_3(1)$
Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=\log(x)$ and $\mathrm dv=\frac{\log(1-x)}x$ to get
\begin{align*}
\int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=\underbrace{\left[\log(x)(-\operatorname{Li}_2(x))\right]_0^1}_{\to0}+\int_0^1\frac{\operatorname{Li}_2(x)}x\mathrm dx\\
&=[\operatorname{Li}_3(x)]_0^1\\
&=\zeta(3)
\end{align*}
A quick look at the series representation of the Trilogarithm verifies the last line.
A: $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,.$$
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.
A: The Beta function and Feynman's trick are another way to go:
$$I=\int_{0}^{1}\frac{\log(x)\log(1-x)}{x}\,dx =\left.\frac{\partial^2}{\partial a \partial b}\int_{0}^{1}x^{a-1}(1-x)^{b}\,dx\,\right|_{\alpha,\beta=0^+}\tag{1} $$
hence:
$$ I = \left.\frac{\partial^2}{\partial a \partial b}\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}\,\right|_{\alpha,\beta=0^+}\tag{2} $$
and by exploiting $\Gamma'(z) = \Gamma(z)\cdot\psi(z)$ we get:
$$ I = -\frac{1}{2}\psi''(2)=\sum_{n\geq 0}\frac{1}{(n+1)^3}=\color{red}{\zeta(3)}\tag{3} $$
as wanted.
A: Note
$$\zeta(3)= \mathrm{Li}_3(1)=\int_0^1\frac{\mathrm{Li}_2(x)}xdx
=\int_0^1\left(- \int_0^x\frac{\ln (1-t)}tdt \right)d(\ln x)
= \int_0^1\frac{\ln x\ln(1-x)}{x}dx
$$
