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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x} =
\int_{1}^{2}{\ln^{3}\pars{x} \over x - 1}\,\dd x =
\int_{1}^{1/2}{\ln^{3}\pars{1/x} \over 1/x - 1}
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
-\int_{1/2}^{1}{\ln^{3}\pars{x} \over x\pars{1 - x}}\,\dd x =
\overbrace{-\int_{1/2}^{1}{\ln^{3}\pars{x} \over x}\,\dd x}
^{\ds{{1 \over 4}\ln^{4}\pars{2}}}\ -\
\int_{1/2}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x
\\[5mm] = &\
{1 \over 4}\ln^{4}\pars{2} -
\bracks{\left.-\ln\pars{1 - x}\ln^{3}\pars{x}\right\vert_{\ 1/2}^{\ 1} -
3\int_{1/2}^{1}\mrm{Li}'_{2}\pars{x}\ln^{2}\pars{x}\,\dd x}
\\[5mm] = &\
-\,{3 \over 4}\ln^{4}\pars{2} -3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} -
6\int_{1/2}^{1}\mrm{Li}'_{3}\pars{x}\ln\pars{x}\,\dd x
\\[5mm] = &\
-\,{3 \over 4}\ln^{4}\pars{2} -
3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} -
6\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} + 6\int_{1/2}^{1}\mrm{Li}'_{4}\pars{x}\,\dd x
\\[5mm] = &\
-\,{3 \over 4}\ln^{4}\pars{2} -
3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} -
6\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2}
\\[2mm] & \phantom{\,\,\,}+\
6\,\mrm{Li}_{4}\pars{1} - 6\mrm{Li}_{4}\pars{1 \over 2}
\end{align}
$\ds{\mrm{Li}_{4}\pars{1} = \zeta\pars{4} = \pi^{4}/90}$ and values of $\ds{\mrm{Li}_{2}\pars{1/2}}$ and
$\ds{\mrm{Li}_{3}\pars{1/2}}$
are already known such that the final result is reduced to:
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x}
\\[5mm] = &\
\bbx{{\pi^{4} \over 15} + {1 \over 4}\,\pi^{2}\ln^{2}\pars{2} -
{1 \over 4}\,\ln^{4}\pars{2} -
6\,\mrm{Li}_{4}\pars{1 \over 2} - {21 \over 4}\,\ln\pars{2}\zeta\pars{3}}
\\[5mm] \approx &\ 0.1425
\end{align}