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Con $\ds{r \equiv {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$
\begin{align}&\color{#c00000}{
\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x}
=\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over \pars{x - r}\pars{x - r^{*}}}\,\dd x
\\[3mm]&=\int_{0}^{1}\ln\pars{\ln\pars{1/x}}
\pars{{1 \over x - r} - {1 \over x - r^{*}}}\,{1 \over r - r^{*}}\,\dd x
={1 \over \Im\pars{r}}\,\Im\int_{0}^{1}
{\ln\pars{\ln\pars{1/x}} \over x - r}\,\dd x
\end{align}
Con $\ds{x \equiv \expo{-t}}$:
\begin{align}&\color{#c00000}{
\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x}
={2\root{3} \over 3}\Im\int_{\infty}^{0}{\ln\pars{t} \over \expo{-t} - r}
\,\pars{-\expo{-t}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r}\int_{0}^{\infty}
{\ln\pars{t}\expo{-t} \over 1 - \expo{-t}/r}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r}
\sum_{n = 1}^{\infty}{1 \over r^{n - 1}}\int_{0}^{\infty}
\ln\pars{t}\expo{-nt}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\sum_{n = 1}^{\infty}r^{-n}\int_{0}^{\infty}t^{\mu}\expo{-nt}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\sum_{n = 1}^{\infty}{r^{-n} \over n^{\mu + 1}}\Gamma\pars{\mu + 1}}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{r^{*}}}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{\expo{-\pi\ic/3}}}
\end{align}
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. Los derivados de la
PolyLogarithm, el respeto de la orden, puede ser evaluado a partir de su representación integral.
También, ver Hurwitz Zeta Función.