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
$$ \begin{align} \int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx &= \frac{19 \pi^{3}}{648} + \frac{1}{2} \left[ - \frac{ \pi^{3}}{27} - \frac{2 \pi }{3} \left(\frac{\pi^{2}}{36} \right) + 2 \left( \frac{5 \pi^{3}}{162} \right) \right] \\ &= \frac{7 \pi^{3}}{216} . \end{align}$$
Note that
$$
\begin{align*}I &= \int_0^1 \log \Gamma(x+1) \; dx = \int_0^1 \log \left(x \Gamma(x) \right)\; dx = \int_0^1 \log(x)dx+\int_0^1 \log\Gamma(x) \; dx \\
&=-1+\int_0^1 \log\Gamma(x) \; dx \tag{1}
\end{align*}$$
On the substitution $x\mapsto 1-x$, this becomes
$$
I=-1+\int_0^1 \log\Gamma(1-x) \; dx \tag{2}
$$
Averaging $(1)$ and $(2)$,
$$
\begin{align*}
I &= -1+\frac{1}{2}\int_0^1 \log\left( \Gamma(x)\Gamma(1-x) \right)\; dx \\
&= -1+\frac{1}{2}\int_0^1 \log \frac{\pi}{\sin(\pi x)}dx \tag{3}\\
&= -1+\frac{1}{2}\log(\pi)-\frac{1}{2}\int_0^1 \log(\sin \pi x) dx \\
&= -1+\frac{1}{2}\log(\pi)-\frac{1}{2\pi}\int_0^\pi \log(\sin x)dx \\
&= -1+\frac{1}{2}\log(\pi)+\frac{1}{2}\log(2) \tag{4}\\
&= \log\sqrt{2\pi}-1
\end{align*}
$$
Equation $(3)$ follows from the reflection formula of the Gamma Function. To get equation $(4)$, I used the well known result:
$$\int_0^\pi \log(\sin x)dx=-\pi \log(2)$$
Best Answer
Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields $$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt $$ $$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) $$ Then $$I(\alpha)=\frac{1}{4} \int\left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) d\alpha= $$ $$I(\alpha)=\left(\ln \Gamma \left(\frac{3 + \alpha}{4}\right)- \ln \Gamma \left(\frac{1 + \alpha}{4}\right)\right)+C\tag1$$ If letting $\alpha=2$, then $$I(2)=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)+C$$ On the other hand, by letting $\alpha=0$ in $(1)$ we get $$C=\ln \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)$$ Thus $$\int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)\Gamma \left(\frac{1}{4}\right)}{\Gamma^2 \left(\frac{3}{4}\right)}\right) $$