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$\ds{\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x:\ {\large ?}}$
\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}
=\int_{1}^{2}{\ln^{2}\pars{x} \over x - 1}\,\dd x
=\int_{1}^{1/2}{\ln^{2}\pars{1/x} \over 1/x - 1}\,\pars{-\,{\dd x \over x^{2}}}
=\int_{1/2}^{1}{\ln^{2}\pars{x} \over x\pars{1 - x}}\,\dd x
\\[3mm]&=\int_{1/2}^{1}{\ln^{2}\pars{x} \over x}\,\dd x
+ \int_{1/2}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x
={1 \over 3}\,\ln^{3}\pars{2}
+\sum_{n = 1}^{\infty}\color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x}
\qquad\qquad\pars{1}
\end{align}
$$
\color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x}
=\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu}\int_{1/2}^{1}x^{\mu}\,\dd x
=\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu}
\bracks{{1 - \pars{1/2}^{\mu + 1} \over \mu + 1}}
$$
$$
\color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x}
=-2\,{\pars{1/2}^{n} \over n^{3}}+ {2 \over n^{3}}
-\ln^{2}\pars{2}\,{\pars{1/2}^{n} \over n}
-2\ln\pars{2}\,{\pars{1/2}^{n} \over n^{2}}
$$
By replacing in $\pars{1}$:
\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}
\\[3mm]&={1 \over 3}\,\ln^{3}\pars{2} -2{\rm Li}_{3}\pars{\half}
+2\zeta\pars{3} - \ln^{2}\pars{2}{\rm Li}_{1}\pars{\half}
-2\ln\pars{2}{\rm Li}_{2}\pars{\half}\tag{2}
\end{align}
You'll find values for the PolyLogarithm Function
$\ds{{\rm Li}_{s}\pars{\half}\,,\ \pars{~s = 1,2,3~}\,,\ }$ in
this page:
\begin{align}
{\rm Li}_{1}\pars{\half} &= \ln\pars{2}
\\[3mm]
{\rm Li}_{2}\pars{\half} &= {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}
\\[3mm]
{\rm Li}_{3}\pars{\half} &= {1 \over 6}\,\ln^{3}\pars{2}- {\pi^{2} \over 12}\,\ln\pars{2}
+{7 \over 8}\,\zeta\pars{3}
\end{align}
With these identities and result $\pars{2}$:
\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}
\\[3mm]&=\color{#00f}{{1 \over 3}\,\ln^{3}\pars{2}} +\
\overbrace{\bracks{\color{#00f}{-\,{1 \over 3}\,\ln^{3}\pars{2}}
+
\color{magenta}{{\pi^{2} \over 6}\,\ln\pars{2}}
{\large -{7 \over 4}\,\zeta\pars{3}}}}^{\ds{-2{\rm Li}_{3}\pars{\half}}}\
+\ {\large 2\zeta\pars{3}}
\\[3mm]&+\
\underbrace{\bracks{\color{#990099}{-\ln^{3}\pars{2}}}}
_{\ds{-\ln^{2}\pars{2}{\rm Li}_{1}\pars{\half}}}\ +\
\underbrace{\bracks{\color{magenta}{-\,{\pi^{2} \over 6}\,\ln\pars{2}}
+\color{#990099}{\ln^{3}\pars{2}}}}_{\ds{-2\ln\pars{2}{\rm Li}_{2}\pars{\half}}}\
=\ \pars{2 - {7 \over 4}}\zeta\pars{3}
\end{align}
$$
\color{#66f}{\large%
\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x = {\zeta\pars{3} \over 4}}
\approx 0.3005
$$