Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ and do a comparision with $\zeta(3)$?

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.

\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}

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}