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\begin{align}
&\bbox[#ffd,10px]{\ds{%
\int_{0}^{1}{\ln^{4}\pars{1 + x}\ln\pars{x} \over x}\,\dd x}}
\,\,\,\stackrel{1 + x\ \mapsto\ x}{=}\,\,\,
\int_{1}^{2}{\ln^{4}\pars{x}\ln\pars{x - 1} \over x - 1}\,\dd x
\\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,&
\int_{1}^{1/2}{\ln^{4}\pars{1/x}\ln\pars{1/x - 1} \over 1/x - 1}
\,\pars{-\,{\dd x \over x^{2}}} =
\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} - \ln^{5}\pars{x} \over
x\pars{1 - x}}\,\dd x
\\[1cm] = &\
\underbrace{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x}
_{\ds{\mc{I}_{1}}}\ -\
\underbrace{\int_{1/2}^{1}{\ln^{5}\pars{x} \over x}\,\dd x}
_{\ds{=\ -\,{\ln^{6}\pars{2} \over 6}}}\ +\
\underbrace{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x}
_{\ds{\mc{I}_{2}}}
\\[1mm] &\
-\int_{1/2}^{1}{\ln^{5}\pars{x} \over 1 - x}\,\dd x
\label{1}\tag{1}
\end{align}
Note that
\begin{align}
\int_{1/2}^{1}{\ln^{5}\pars{x} \over 1 - x}\,\dd x & =
-\ln^{6}\pars{2} +
5\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x =
-\ln^{6}\pars{2} + 5\,\mc{I}_{1}
\end{align}
such that \eqref{1} becomes
\begin{equation}
\bbx{\bbox[#ffd,10px]{\ds{%
\int_{0}^{1}{\ln^{4}\pars{1 + x}\ln\pars{x} \over x}\,\dd x}} =
{7\ln^{6}\pars{2} \over 6} - 4\,\mc{I}_{1} + \mc{I}_{2}}\label{2}\tag{2}
\end{equation}
$\ds{\Huge\mc{I}_{1}:\ ?}$.
\begin{align}
\mc{I}_{1} & \equiv
\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x =
-\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{4}\pars{x}\,\dd x
\\[5mm] & =
\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} +
4\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\ln^{3}\pars{x}\,\dd x
\\[5mm] & =
\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} +
4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} -
12\int_{1/2}^{1}\mrm{Li}_{4}'\pars{x}\ln^{2}\pars{x}\,\dd x
\\[5mm] & =
\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} +
4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} +
12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} +
24\int_{1/2}^{1}\mrm{Li}_{5}'\pars{x}\ln\pars{x}\,\dd x
\\[1cm] & =
\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} +
4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} +
12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} +
24\,\mrm{Li}_{5}\pars{1 \over 2}\ln\pars{2}
\\[1mm] & - 24\int_{1/2}^{1}\mrm{Li}_{6}'\pars{x}\,\dd x
\\[1cm] & =
\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} +
4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} +
12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} +
24\,\mrm{Li}_{5}\pars{1 \over 2}\ln\pars{2}
\\[1mm] & -24\,\mrm{Li}_{6}\pars{1} + 24\,\mrm{Li}_{6}\pars{1 \over 2}
\end{align}
Note that
$\ds{\mrm{Li}_{2}\pars{1 \over 2} = {\pi^{2} \over 12} -
{\ln^{2}\pars{2} \over 2}}$ and
$\ds{\mrm{Li}_{3}\pars{1 \over 2} = {\ln^{3}\pars{2} \over 6} -
{\pi^{2}\ln\pars{2} \over 12} + {7\zeta\pars{3} \over 8}}$. Moreover,
$\ds{\vphantom{\Huge A}\mrm{Li}_{6}\pars{1} = \zeta\pars{6} =
{\pi^{6} \over 945}}$.
Then,
$$
\begin{array}{|rcl|}\hline \mbox{}&&\\
\ds{\quad\mc{I}_{1}} & \ds{\equiv} &
\ds{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x}
\\[5mm] & \ds{=} &
\ds{-\,{8\pi^{6} \over 315} - {\pi^{2}\ln^{4}\pars{2} \over 4} +
{\ln^{6}\pars{2} \over 6} + 12\ln^{2}\pars{2}\,\mrm{Li}_{4}\pars{1 \over 2} +
24\ln\pars{2}\,\mrm{Li}_{5}\pars{1 \over 2}\quad}
\\[1mm] &&
\ds{%
+ 24\,\mrm{Li}_{6}\pars{1 \over 2} + {7\ln^{3}\pars{2}\,\zeta\pars{3} \over 2}
\approx -0.0269}
\\ \mbox{}&&
\\ \hline
\end{array}
$$
$\ds{\Huge\mc{I}_{2}:\ ?}$. This one isn't trivial at all. An attempt is given by
\begin{align}
\mc{I}_{2} & \equiv
\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,
\int_{0}^{1/2}{\ln^{4}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\\[5mm] & =
\left.\partiald[4]{}{\mu}\partiald{}{\nu}
\int_{0}^{1/2}\bracks{\pars{1 - x}^{\mu} - 1}x^{\nu - 1}
\,\dd x\,\right\vert_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}}
\\[5mm] & =
\partiald[4]{}{\mu}\partiald{}{\nu}
\bracks{\int_{0}^{1/2}\pars{1 - x}^{\mu}\,x^{\nu - 1}
\,\dd x - {1 \over 2^{\nu}\nu}}_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}}
\\[5mm] & =
\partiald[4]{}{\mu}\partiald{}{\nu}
\bracks{\mrm{B}\pars{{1 \over 2},\nu,1 + \mu} - {1 \over 2^{\nu}\nu}}_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}}
\end{align}
$\ds{\mrm{B}}$ is the Incomplete Beta Function. It'll continue$\ldots$