How does one verify that $\int_{0}^{1}x\ln[{\ln{x}}\ln(1-x)]\mathrm dx=-\gamma?$ Consider $(1)$
How does one verify that

$$\int_{0}^{1}x\ln[\color{red}{\ln{x}}\ln(1-x)]\mathrm dx=-\gamma\tag1$$

An attempt:
$$\int_{0}^{1}x\ln{(\ln{x})}\mathrm dx+\int_{0}^{1}x\ln{(\ln{(1-x))}}\mathrm dx\tag2$$
$u=\ln{x}$ then $xdu=dx$
$$\int_{0}^{\infty}e^{-2u}\ln{(2u)}\mathrm du+\int_{0}^{\infty}e^{-2u}\ln{(1-e^u)}\mathrm du=I_1+I_2\tag3$$
Applying integration by parts to $I_1$
$$I_1={-e^{-2u}\ln{(2u)}\over 2}|_{0}^{\infty}+{1\over 2}\int_{0}^{\infty}{1\over u}\cdot e^{-2u}\mathrm du\tag4$$
$(4)$ seem to be divergent
How else can we tackle $(1)?$
 A: There's a subtle problem with your split: the terms you've defined, e.g. $x\ln(\ln(x))$, aren't actually defined (over the reals) on the domain $(0,1)$ because $\ln(x)$ is negative on this range!  Instead, you have to split things as $\displaystyle\int_0^1x\ln(-\ln(x))dx+\int_0^1x\ln(-\ln(1-x))dx$; this works because the minus signs internally cancel when the terms are multiplied!
But now we can substitute $u=1-x$ into the second term: minus signs from $du=-dx$ and from switching the order of integration from $\int_1^0$ back to $\int_0^1$ cancel, leaving the second integral as $\displaystyle\int_0^1(1-u)\ln(-\ln(u))du$.  And then we can add this to the first term of the split to get $\displaystyle\int_0^1\ln(-\ln(x))dx$.  But this formula (more often written in the form $\displaystyle\int_0^1\ln(\ln(\frac1x))dx$) is one of the classical expressions for $-\gamma$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,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}}}
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\begin{align}
&\int_{0}^{1}x\ln\pars{\ln\pars{x}\ln\pars{1 - x}}\,\dd x =
\int_{0}^{1}x\ln\pars{-\ln\pars{x}}\,\dd x +
\int_{0}^{1}x\ln\pars{-\ln\pars{1 - x}}\,\dd x
\\[5mm] = &\
\int_{0}^{1}x\ln\pars{-\ln\pars{x}}\,\dd x +
\int_{0}^{1}\pars{1 - x}\ln\pars{-\ln\pars{x}}\,\dd x = \int_{0}^{1}\ln\pars{-\ln\pars{x}}\,\dd x
\\[5mm] \stackrel{x\ =\ \exp\pars{-t}}{=} &\
\int_{0}^{\infty}\ln\pars{t}\expo{-t}\,\dd t = \bbx{\ds{-\,\gamma}}
\end{align}
A: Letting $x \mapsto 1-x$,
$$
I = \int_{0}^{1} \ln [\ln (1-x) \ln x] d x-I
$$
we have $$
\begin{aligned}
I &=\frac{1}{2} \int_{0}^{1} \ln [\ln (1-x) \ln x] d x \\
&=\frac{1}{2}\left[\int_{0}^{1} \ln [-\ln (1-x)]+\ln (-\ln x)d x\right] \\
&=\frac{1}{2}\left[\int_{0}^{1} \ln (-\ln x)d x+\int_{0}^{1} \ln (-\ln x) d x\right] \\
&=\int_{0}^{1} \ln (-\ln x) d x \\&= \int_{0}^{\infty} e^{-x} \ln x d x\\
&=-\gamma
\end{aligned}
$$
