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With $\ds{r \equiv {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$
\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
\\[3mm] & =
{1 \over \Im\pars{r}}\,\Im\int_{0}^{1}
{\ln\pars{\ln\pars{1/x}} \over x - r}\,\dd x
\end{align}
With $\ds{x \equiv \expo{-t}}$:
\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}
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. Derivatives of the
PolyLogarithm, respect of the order, can be evaluated from its integral representation.
Also, see Hurwitz Zeta Function.
You can use the identity given by the Euler Beta function
$$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
to state:
$$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$
and by switching the series and the integral:
$$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$
$$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$
$$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$
Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity
$$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$
that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.
As a matter of fact, both approaches lead to an answer.
The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity:
$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$
while the integral approach, as Achille Hui pointed out in the comments, leads to:
$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$
Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.
Update 14-06-2016. I just discovered that this problem can also be solved by computing
$$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$
where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.
Best Answer
If you don't mind, I would like to present an alternative approach that makes use of the fact that $$\int^\infty_0\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin{p\pi}}$$ Simply factorise the denominator and decompose the integrand into partial fractions. \begin{align} \int^\infty_0\frac{x^a}{x^2+2(\cos{b})x+1}dx &=\int^\infty_0\frac{x^a}{(x+e^{ib})(x+e^{-ib})}dx\\ &=\frac{1}{-e^{ib}+e^{-ib}}\int^\infty_0\frac{x^a}{e^{ib}+x}dx+\frac{1}{-e^{-ib}+e^{ib}}\int^\infty_0\frac{x^a}{e^{-ib}+x}dx\\ &=\frac{1}{-2i\sin{b}}\int^\infty_0\frac{(e^{ib}u)^a}{1+u}du+\frac{1}{2i\sin{b}}\int^\infty_0\frac{(e^{-ib}u)^a}{1+u}du\\ &=\frac{e^{iab}}{-2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}+\frac{e^{-iab}}{2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}\\ &=\frac{\pi}{\sin \pi a\sin{b}}\left(\frac{e^{iab}-e^{-iab}}{2i}\right)\\ &=\frac{\pi\sin{ab}}{\sin{\pi a}\sin{b}} \end{align}