$$J=\int_0^{\pi/2}\ln\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx\overset{2x=t}=\frac12 \int_0^\pi \ln\left(\frac{1+\frac12\sin t}{1-\frac12\sin t}\right)\mathrm dt=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\frac12\sin x}{1-\frac12\sin x }\right)\mathrm dx$$
Now let's consider the following integral:
$$I(a)=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\sin a\sin x}{1-\sin a\sin x}\right)dx\Rightarrow I'(a)=2\int_0^\frac{\pi}{2} \frac{\cos a\sin x}{1-\sin^2a\sin^2 x}dx$$
$$=\frac{2\cos a}{\sin^2a}\int_0^\frac{\pi}{2} \frac{\sin x}{\cos^2x +\cot^2 a}dx=\frac{2}{\sin a}\,\arctan\left(x\tan a\right)\bigg|_0^1=\frac{2a}{\sin a}$$
$$I(0)=0 \Rightarrow J=I\left(\frac{\pi}{6}\right)=2\int_0^\frac{\pi}{6}\frac{x}{\sin x}dx$$
$$=2\int_0^{\frac{\pi}{6}} x \left(\ln\left(\tan \frac{x}{2}\right)\right)'dx=2x \ln\left(\tan \frac{x}{2}\right)\bigg|_0^{\frac{\pi}{6}} -2{\int_0^{\frac{\pi}{6}} \ln\left(\tan \frac{x}{2}\right)dx}=$$
$$\overset{\frac{x}{2}=t}=\frac{\pi}{3}\ln(2-\sqrt 3) -4\int_0^\frac{\pi}{12}\ln (\tan t)dt=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G$$ $G$ is the Catalan's constant and for the last integral see here.
Also note that there's a small mistake. After integrating by parts you should have: $$2I=\frac{\pi^2}{4\sqrt 3}- \int_0^\infty\frac{(x^2-1)\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{4\sqrt 3}-\frac12\underbrace{\int_0^\infty \ln\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{dx}{1+x^2}}_{=J}$$
Here is one approach. As a warning, the final answer I find is not pretty.
Let
$$I = \int_0^1 \frac{\tan^{-1} x}{x^2 -x - 1} \, dx.$$
Start by using a self-similar substitution of
$$x = \frac{1 - u}{1 + u}, \,\, dx = -\frac{2}{(1 + u)^2} \, du.$$
So , after having reverted the dummy variable $u$ back to $x$, we have
$$I = 2 \int_0^1 \frac{\tan^{-1} \left (\frac{1 - x}{1 + x} \right )}{x^2 - 4x - 1} \, dx.$$
Noting that for $0 < x < 1$
$$\tan^{-1} \left (\frac{1 - x}{1 + x} \right ) = \frac{\pi}{4} - \tan^{-1} x,$$
then
\begin{align}
I &= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 - 4x - 1} - 2 \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx\\
&= -\frac{\pi}{2 \sqrt{5}} \coth^{-1} \left (\frac{3}{\sqrt{5}} \right ) - 2 J,
\end{align}
where
$$J = \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx.$$
To find $J$ we begin by noting that $\tan^{-1} x = \operatorname{Im} \ln (1 + ix)$. Thus
$$J = \operatorname{Im} \int_0^1 \frac{\ln (1 + ix)}{x^2 - 4x - 1} \, dx.$$
Making a substitution of $t = 1 + ix$ we have
$$J = - \operatorname{Re} \int_1^{1+i} \frac{\ln t}{(t - \alpha)(t - \beta)} \, dt,$$
where $\alpha = 1 + i(2 - \sqrt{5})$ and $\beta = 1 + i(2 + \sqrt{5})$. After performing a partial fraction decomposition we are left with
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\int_1^{1 + i} \frac{\ln t}{\alpha - t} \, dt - \int_1^{1 + i} \frac{\ln t}{\beta - t} \, dt \right ].$$
Now, as
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ),$$
(for a proof of this result see the appendix below), one has
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\ln (1 + i) \ln \left (\frac{\alpha}{\beta} \cdot \frac{\beta - 1 - i}{\alpha - i - i} \right ) + \operatorname{Li}_2 \left (\frac{1}{\alpha} \right ) - \operatorname{Li}_2 \left (\frac{1}{\beta} \right ) + \operatorname{Li}_2 \left (\frac{1 + i}{\beta} \right ) - \operatorname{Li}_2 \left (\frac{1 + i}{\alpha} \right ) \right ]$$
or after performing a huge amount of algebra
\begin{align}
J &= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\operatorname{Li}_2 \left (\frac{1}{2} + \frac{1}{\sqrt{5}}+ \frac{i}{2 \sqrt{5}} \right ) - \operatorname{Li}_2 \left (\frac{1}{2} - \frac{1}{\sqrt{5}} - \frac{i}{2 \sqrt{5}} \right ) \right.\\
& \quad+ \left. \operatorname{Li}_2 \left (\frac{1}{2} -\frac{1}{2 \sqrt{5}} - i \left (\frac{3}{2 \sqrt{5}} - \frac{1}{2} \right ) \right ) - \operatorname{Li}_2 \left (\frac{1}{2} +\frac{1}{2 \sqrt{5}} + i \left (\frac{3}{2 \sqrt{5}} + \frac{1}{2} \right ) \right ) \right ]\\
&= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \frak{w},
\end{align}
where $\frak{w}$ is the term containing the four dilogarithms with complex arguments. Thus
$$\int_0^1 \frac{\tan^{-1} x}{x^2 - x - 1} \, dx = -\frac{\pi}{4 \sqrt{5}} \left (\ln 2 + \sinh^{-1} (2) \right ) - \frac{1}{\sqrt{5}} \operatorname{Im} \frak{w}.$$
Note that as $\operatorname{Im} {\frak{w}} = -0.8363170651979\ldots$ we see that $I \approx -0.376513$.
Appendix
Proof of
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C$$
Setting $t = x/z, dt = dx/z$, we have
\begin{align}
\int \frac{\ln x}{z - x} \, dx &= \int \frac{\ln (zt)}{1 - t} \, dt\\
&= -\ln (1 - t) \ln (zt) + \int \frac{\ln (1 - t)}{t} \, dt \qquad \text{(by parts)}\\
&= -\ln (1 - t) \ln (zt) - \operatorname{Li}_2 (t) + C\\
&= - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C,
\end{align}
as required to show.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x^{2}{\ln\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x} \\[5mm] = &\ \left. \Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\bracks{-\ic\ln\pars{z}}^{\, 2}\,{\ln\pars{\bracks{z - 1/z}/\bracks{2\ic}} \over \pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left. 2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln^{2}\pars{z} \ln\pars{{1 - z^{2} \over 2z}\,\ic}\,{\dd z \over 1 - z^{2}} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = & -2\,\Re\int_{1}^{0}\bracks{\ln\pars{y} + {\pi \over 2}\,\ic}^{2} \ln\pars{1 + y^{2} \over 2y}\,{\ic\,\dd y \over 1 + y^{2}} \\[2mm] &\ -\! 2\,\Re\int_{0}^{1}\ln^{2}\pars{x} \ln\pars{{1 - x^{2} \over 2x}\,\ic}\,{\dd x \over 1 - x^{2}} \\[5mm] = & -2\pi\int_{0}^{1}\ln\pars{y}\ln\pars{1 + y^{2} \over 2y} \,{\dd y \over 1 + y^{2}} \\[2mm] & \,\, -2\int_{0}^{1}\ln^{2}\pars{x} \ln\pars{1 - x^{2} \over 2x}\,{\dd x \over 1 - x^{2}} \\[5mm] = & -2\pi\ \overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y^{2}} \over 1 + y^{2}}\,\dd y}^{\ds{I_{1}}}\ +\ 2\pi\ln\pars{2}\ \overbrace{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{2}}} \\[2mm] &\ +2\pi\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{y} \over 1 + y^{2}}\,\dd y} _{\ds{I_{3}}} \\[2mm] & \,\, -2\ \overbrace{\int_{0}^{1}{\ln^{2}\pars{x} \ln\pars{1 - x^{2}} \over 1 - x^{2}}\,\dd x}^{\ds{I_{4}}}\ +\ 2\ln\pars{2}\ \overbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}} \,\dd x}^{\ds{I_{5}}} \\[2mm] &\ + 2\ \underbrace{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x^{2}}\,\dd x} _{\ds{I_{6}}} \\[5mm] = &\ -2\pi I_{1} + 2\pi\ln\pars{2}I_{2} + 2\pi I_{3} - 2I_{4} + 2\ln\pars{2}I_{5} + 2I_{6}\label{1}\tag{1} \end{align}
\begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\left\{\begin{array}{rcl} \ds{\LARGE\bullet} && \ds{I_{2}, I_{3}, I_{5}}\ \mbox{and}\ \ds{I_{6}}\ \mbox{are rather trivial ones or/and amenable to standard} \\ && \mbox{techniques}\ (~\mbox{IBP, Polylogarithms, rescaling, etc$\ldots$}~) \\[1mm] \ds{\LARGE\bullet} && \ds{I_{4}}\ \mbox{can be evaluated via the Beta Function, after the rescaling}\ \ds{x^{2} \mapsto x}. \\[1mm] \ds{\LARGE\bullet} && \mbox{After the rescaling}\ \ds{y^{2} \mapsto y},\ \ds{I_{1}}\ \mbox{can be written as a sum that involves} \\ && \mbox{the}\ harmonic\ number\ \mbox{because}\ \ds{{\ln\pars{1 + x} \over 1 + x} = -\sum_{k = 1}^{\infty}H_{k}\,\pars{-1}^{k}x^{k}}. \\ && \mbox{It turns out that}\ \ds{I_{1} = \sum_{k = 1}^{\infty}\pars{-1}^{k}\,{H_{k} \over \pars{2k + 1}^{2}}} \\[3mm] && \mbox{Indeed,}\ \ds{I_{1}}\ \mbox{was}\ \underline{evaluated}\ \mbox{in a previous post by user}\ {\tt @user97357329}. \\ && \mbox{See the link at the very end.} \\ && \S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S\S \\[2mm] \ds{I_{1}} & \ds{=} & \ds{-\,{\pi^{3} \over 64} - \ln\pars{2}G - {\pi\ln^{2}\pars{2} \over 16} + 2\,\Im\mrm{Li}_{3}\pars{1 + \ic \over 2}} \\[2mm] \ds{I_{2}} & \ds{=} & \ds{-G\,\qquad\pars{~G:\ Catalan\ Constant~}} \\[2mm] \ds{I_{3}} & \ds{=} & \ds{\phantom{-}{\pi^{3} \over 16}} \\[2mm] \ds{I_{4}} & \ds{=} & \ds{-\,{\pi^{4} \over 32} + {7\ln\pars{2}\zeta\pars{3} \over 2}} \\[2mm] \ds{I_{5}} & \ds{=} & \ds{\phantom{-}{7\zeta\pars{3} \over 4}} \\[2mm] \ds{I_{6}} & \ds{=} & \ds{-\,{\pi^{4} \over 16}} \end{array}\right. \label{2}\tag{2} \end{equation}
Finally, with (\ref{1}) and (\ref{2}): \begin{align} I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x^{2}{\ln\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x} \\[5mm] & = \bbx{{3\pi^{4} \over 32} + {\pi^{2}\ln^{2}\pars{2} \over 8} - 4\pi\,\Im\mrm{Li}_{3}\pars{1 + \ic \over 2} - {7\ln\pars{2}\zeta\pars{3} \over 2}} \\ & \end{align}
Thanks to user ${\tt @Ali Shather}$ who calls my attention to a link where $\ds{I_{1}}$ is evaluated.