By the substitution $x = e^{-t}$, we find that
$$\begin{align*}
\int_{0}^{1} \frac{(\log (1/x))^s}{1+x^2} \; dx
&= \int_{0}^{\infty} \frac{t^s e^{-t}}{1 + e^{-2t}} \; dt \\
&= \int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n t^s e^{-(2n+1)t} \; dt \\
&= \sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \\
&= \Gamma(s+1)L(s+1, \chi_4),
\end{align*}$$
where $L(s, \chi_4)$ is the Dirichlet L-function of the unique non-principal character $\chi_4$ to the modulus 4. Often it is denoted as $\beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \psi_0(1) \beta(1) + \beta'(1),$$
and the problem reduces to find the value of $\beta(1)$ and $\beta'(1)$. Note that $\beta(1) = \frac{\pi}{4}$ is just the Gregory series. For $\beta'(1)$, we first notice that the following functional equation holds.
$$ \beta(s)=\left(\frac{\pi}{2}\right)^{s-1} \Gamma(1-s) \cos \left( \frac{\pi s}{2} \right)\,\beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $\beta'(0)$. For $0 < s$, we have
$$\begin{align*}
-\beta'(s)
&= \sum_{n=1}^{\infty} \left[ \frac{\log(4n+1)}{(4n+1)^s} - \frac{\log(4n-1)}{(4n-1)^s} \right] \\
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right] + 2^{-2s-1}\zeta(s+1) \\
& \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n+1)^s} - \frac{1}{(4n)^s} \right) \log (4n+1) \\
& \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n)^s} - \frac{1}{(4n-1)^s} \right) \log (4n-1) \\
& =: A(s) + 2^{-2s-1}\zeta(s+1) + B(s) + C(s).
\end{align*}$$
We first estimate $B(s)$. As $n \to \infty$, we have
$$ \log \left( \frac{4n}{4n+1} \right) = -\frac{1}{4n} + O\left( \frac{1}{n^2} \right), \quad \log \left( \frac{4n}{4n-1} \right) = \frac{1}{4n} + O\left( \frac{1}{n^2} \right). $$
Thus when $s \to 0$,
$$\begin{align*}
B(s)
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \exp\left( s \log \left( \frac{4n}{4n+1} \right) \right) - 1 \right] \left[ \log (4n) - \log \left(\frac{4n}{4n+1} \right) \right] \\
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ - \frac{s}{4n} + O \left(\frac{s^2}{n^2} \right) \right] \left[ \log (4n) + O \left(\frac{1}{n} \right) \right] \\
&= -s 2^{-2s-2} \sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \log (4n) + O(s) \\
&= s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).
\end{align*}$$
Similar consideration also shows that
$$ C(s) = s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).$$
Thus we have
$$ 2^{-2s-1}\zeta(s+1) + B(s) + C(s) = 2^{-2s-1} \left[ \zeta(s+1) + s \zeta'(s+1) - s \zeta(s+1) \log 4 \right] + O(s). $$
But since
$$\zeta(1+s) = \frac{1}{s} + \gamma + O(s),$$
we have
$$ \lim_{s\downarrow 0} \left( 2^{-2s-1}\zeta(s+1) + B(s) + C(s) \right) = \frac{\gamma}{2} - \log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ \lim_{s\downarrow 0} A(s) = \sum_{n=1}^{\infty} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$\begin{align*}
e^{L}
& \stackrel{N\to\infty}{\sim} \prod_{n=1}^{N} \left( \frac{4n+1}{4n-1} \right) e^{-1/2n}
\sim \frac{e^{-\gamma/2}}{\sqrt{N}} \prod_{n=1}^{N} \left( \frac{n+(1/4)}{n-(1/4)} \right) \\
& \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\Gamma\left(N+\frac{5}{4}\right)}{\Gamma\left(N+\frac{3}{4}\right)}
\sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\left( \frac{N + (5/4)}{e} \right)^{N+\frac{5}{4}}}{\left( \frac{N + (3/4)}{e} \right)^{N+\frac{3}{4}}} \\
& \sim e^{-\gamma/2} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}
= 4 e^{-\gamma/2} \frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)^2},
\end{align*}$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-\beta'(0) = \log (2 \pi \sqrt{2}) - 2 \log \Gamma\left(\frac{1}{4}\right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ \frac{\beta'(s)}{\beta(s)} = \log\left(\frac{\pi}{2}\right) - \psi_0 (1-s) - \frac{\pi}{2} \tan \left( \frac{\pi s}{2} \right) - \frac{\beta'(1-s)}{\beta(1-s)}. $$
Taking $s = 0$, we have
$$ \frac{\beta'(0)}{\beta(0)} = \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(1)}{\beta(1)} \quad \Longrightarrow \quad \beta'(1) = \beta(1) \left[ \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(0)}{\beta(0)} \right]. $$
But again by the functional equation, we have $\beta(0) = \frac{1}{2}$. Therefore
$$ \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
and hence
$$ \int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \frac{\pi}{4} \left[ 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
which is identical to the proposed answer.
Another way to show $$\int_{0}^{1} \log \Gamma(x+ \alpha) \, dx = \int_{0}^{1} \log \Gamma(x) \, dx + \alpha \log \alpha - \alpha $$
is to rewrite the integral as
$$ \begin{align} \int_{0}^{1} \log \Gamma (x+\alpha) \, dx &= \int_{\alpha}^{\alpha+1} \log \Gamma(u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{1}^{\alpha+1} \log \Gamma (u) \, du - \int_{0}^{\alpha} \log \Gamma (u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{0}^{\alpha} \log \Gamma (w+1) \, dw - \int_{0}^{\alpha} \log \Gamma (u) \, du \end{align}$$
and then combine the 2nd and 3rd integrals and use the functional equation $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$
Best Answer
Here's another approach for evaluating the one without the $x$ in front.
First notice that it's equivalent to showing that $$\int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx = \frac{7 \pi^{3}}{216}. $$
Using the principal branch of the logarithm and assuming that $0 < x < \pi$, we have $$ \begin{align} \log(1-e^{2ix}) &= \log (e^{-ix}-e^{ix}) + \log(e^{ix}) \\ &= \log(-2i \sin x) + ix \\ &= \log(2 \sin x) - \frac{i \pi}{2} + ix. \end{align}$$
Squaring both sides and integrating, $$\int_{0}^{\pi /6} \left(\log(2 \sin x) - \frac{i \pi}{2} + ix \right)^{2} \ dx = \int_{0}^{\pi /6} \log^{2} (1-e^{2ix}) \ dx . $$
Then equating the real parts on both sides of the equation, we get
$$\begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \int_{0}^{\pi/6} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi /6} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{19 \pi^{3}}{648} +\text{Re} \int_{C} \log^{2}(1-z) \frac{dz}{2iz} \\ &=\frac{19 \pi^{3}}{648} + \frac{1}{2} \ \text{Im} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz \end{align}$$
where $C$ is the portion of the unit circle from $z=1$ to $z=e^{ \pi i /3}$.
But since $\frac{\log^{2}(1-z)}{z}$ is analytic for $\text{Re}(z) <1$,
$$ \begin{align} \int_{C} \frac{\log^{2}(1-z)}{z} \ dz &= \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz . \end{align} $$
Then integrating by parts twice, we get
$$ \begin{align} \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log^{2}(1-z)}{z} \ dz &= \text{Im} \ \log^{2}(1-z) \log(z) \Bigg|^{e^{\pi i /3}}_{1} + 2 \ \text{Im} \int_{1}^{e^{\pi i /3}} \frac{\log(1-z) \log (z)}{1-z} \ dz \\ &= \text{Im} \ \log^{2}(e^{-\pi i /3}) \log(e^{\pi i /3}) + 2 \ \text{Im} \ \log(1-z) \text{Li}_{2}(1-z) \Bigg|^{e^{\pi i / 3}}_{1} \\ &+ 2 \ \text{Im} \int_{1}^{e^{\pi i / 3}} \frac{\text{Li}_{2}(1-z)}{1-z} \ dz \\ &=- \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(1-z) \Bigg|^{e^{\pi i/3}}_{1} \\ &= - \frac{ \pi^3}{27} - \frac{2 \pi }{3}\text{Im} \ i \ \text{Li}_{2} (e^{- \pi i /3}) - 2 \ \text{Im} \ \text{Li}_{3}(e^{ -\pi i /3}) \\ &= - \frac{\pi^3}{27} - \frac{2 \pi }{3} \sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} +2 \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^3}. \end{align}$$
Integrating both sides of the Fourier series $$\sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k} = \frac{\pi - \theta}{2} \ , \ 0 < \theta < 2 \pi$$
we get
$$\sum_{n=1}^{\infty} \frac{\cos (k \theta)}{k^{2}} = \frac{\theta^{2}}{4} - \frac{\pi \theta}{2} + \frac{\pi^{2}}{6} .$$
And integrating a second time, $$ \sum_{n=1}^{\infty} \frac{\sin (k \theta)}{k^{3}} = \frac{\theta^{3}}{12} - \frac{\pi \theta^{2}}{4} + \frac{\pi^{2} \theta}{6}.$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{\cos (n \pi /3)}{n^{2}} = \frac{\pi^{2}}{36} $$
and $$ \sum_{n=1}^{\infty} \frac{\sin (n \pi /3)}{n^{3}} = \frac{5 \pi^{3}}{162}. $$
So finally we have
$$ \begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \frac{19 \pi^{3}}{648} + \frac{1}{2} \left[ - \frac{ \pi^{3}}{27} - \frac{2 \pi }{3} \left(\frac{\pi^{2}}{36} \right) + 2 \left( \frac{5 \pi^{3}}{162} \right) \right] \\ &= \frac{7 \pi^{3}}{216} . \end{align}$$