We will look into the integral
\begin{align*}
I = &\int_{0}^{1}\arcsin^4 x\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx
\end{align*} taking the @nospoon's novel approach presented here. Using the MacLaurin series of $\arcsin^4 x$
$$
\arcsin^4 x =\frac 3 2 \sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}x^{2n}
$$
and the fact that
$$
\small\operatorname{B}(n+\tfrac 1 2,\tfrac 1 2) = \int_0^1 x^{n-1/2}(1-x)^{-1/2}\ \mathrm dx = 2\int_0^{\frac\pi 2} \sin^{2n}\theta\ \mathrm d\theta = \frac{\pi}{4^n}{2n \choose n},\tag{$\small x\mapsto \sin^2\theta$}
$$
\begin{align*}\small
\psi(n+\tfrac 12 ) -\psi(n+1) =&\small \sum_{k=1}^\infty \frac 1{\scriptsize k+n} - \frac 1{\scriptsize k+n-\tfrac 1 2} \\
=&\small\sum_{k=1}^\infty \left(\frac 1{\scriptsize k} - \frac 1{\scriptsize k-\tfrac 1 2}\right)-\sum_{k=1}^n\frac 1 {\scriptsize k} + \sum_{k=1}^n\frac 1{\scriptsize k-\tfrac 1 2}\\
=&\small-2\ln 2 -H_n +2(H_{2n}-\tfrac 1 2H_n)\\
=&\small 2(H_{2n}-H_n-\ln 2),
\end{align*}
\begin{align*}
\Longrightarrow \ {\int_{ 0}^{1 }x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx} = & \frac 1 4\int_{0 }^{1 }x^{n-1/2} { \ln x \over \sqrt{1-x}}\ \mathrm dx\tag{$\small x^2\mapsto x$}\\
=& \frac 1 4 \left[\frac{\partial }{\partial x}\operatorname{B}(x,y) \right]_{x=n+1/2,y=1/2}\\
=&\frac 1 4\Big[ \operatorname{B}(x,y)\big[\psi(x) -\psi(x+y) \big]\Big]_{x=n+1/2,y=1/2}\\
=& \frac 1 4 \operatorname{B}(n+\tfrac 1 2,\tfrac 1 2)\big[\psi(n+\tfrac 12 ) -\psi(n+1) \big]\\
=& \frac{\pi}2\frac{{2n \choose n}}{4^{n}} \left(H_{2n} - H_n -\ln 2\right),
\end{align*} where $\operatorname{B}(x,y)$ and $\psi(x)$ are the Beta and digamma function, respectively, we have
\begin{align*}
I = &\frac 3 2\sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}\int_{0}^{1}x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \\
=&\frac {3\pi}4 \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}\left(H_{2n} - H_n -\ln 2\right) \\
=&\frac {3\pi}4\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{2n}}{n^2}-\frac {3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}}_{=-2\zeta(5) +2\zeta(2)\zeta(3)}-\frac {3\pi\ln 2}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}}_{=\frac{3}4 \zeta(4)}\\
=&\frac{3\pi}{4} \sum_{n=1}^\infty \frac{H^{(2)}_{n}H_{2n}}{n^2} -\frac{3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H_{2n}}{n^4}}_{=\frac{37}{4}\zeta(5)-4\zeta(2)\zeta(3)} +\frac{3\pi}2 \zeta(5) -\frac{\pi^3}4\zeta(3) -\frac{\pi^5\ln 2}{160}\\
=&\boxed{3\pi S -\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}}
\end{align*} where $S = \sum_{n=1}^\infty \frac{H_{2n}H^{(2)}_{n}}{4n^2}$ is the sum in question, and the known values of several Euler sums
$$
\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}=-2\zeta(5) +2\zeta(2)\zeta(3),\tag{1}
$$
$$\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}=\frac{7}4 \zeta(4),\tag{2}
$$
\begin{align*}\sum_{n=1}^\infty \frac{H_{2n}}{n^4} =& 8\sum_{n=1}^\infty \frac{H_{n}}{n^4}-8\sum_{n=1}^\infty \frac{(-1)^{n-1} H_{n}}{n^4}\\
=&8\big(3\zeta(5)-\zeta(2)\zeta(3)\big)-8\left(\frac{59}{32}\zeta(5)-\frac 1 2\zeta(2)\zeta(3)\right)\\
=&\frac{37}4\zeta(5) - 4\zeta(2)\zeta(3)\tag{3}
\end{align*} are used.
Note: $(1)$ is in @nospoon's answer here, $(2)$ can be found here, and for $(3)$ you can see Euler's formula and here.
Evaluation of $I$: By making substitution $x = \sin \theta$ and using the Fourier series of
$$
\ln (\sin\theta) = -\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k},
$$ we get
\begin{align*}
I =& \int_{0}^{\frac\pi 2} \theta^4 \ln(\sin\theta)\ \mathrm d\theta\\
=&\int_{0}^{\frac\pi 2} \theta^4\left(-\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}\right)\ \mathrm d\theta\\
=& -\ln 2\int_0^{\frac \pi 2}\theta^4\ \mathrm d\theta-\sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_{0}^{\frac\pi 2}\theta^4 \cos(2k \theta) \ \mathrm d\theta}_{\text{IBP}\times 4}\\
=& -\frac{\pi^5\ln 2}{160}-\sum_{k=1}^\infty \frac{1}{k}\cdot\left(-\frac{\pi^3}{8}\frac{(-1)^{k-1}}{k^2} +\frac{3\pi}{4}\frac{(-1)^{k-1}}{k^4}\right)\\
=&-\frac{\pi^5\ln 2}{160}+\frac{\pi^3}8\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^3}}_{=\frac 3 4 \zeta(3)} - \frac{3\pi}4\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}}_{=\frac{15}{16}\zeta(5)}\\
=&\boxed{-\frac{\pi^5\ln 2}{160}+\frac{3\pi^3}{32}\zeta(3) -\frac{45\pi}{64}\zeta(5).}
\end{align*}
Combining these, we get the equation
$$
3\pi S-\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}=-\frac{\pi^5\ln 2}{160} +\frac{3\pi^3}{32}\zeta(3)-\frac{45\pi}{64}\zeta(5),
$$hence it follows
$$
\boxed{S = \frac{101}{64}\zeta(5) -\frac{5\pi^2}{96}\zeta(3).}
$$
Addendum: By considering MacLaurin series of
\begin{align*}
\ln(1-x)\ln(1+x)
=&-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)x^{2k}
\end{align*} and
\begin{align*}
\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}
=& \frac{\partial }{\partial k}\left[-\frac{H_k}{k}\right]\\
=& \int_0^1 x^{k-1}\ln x\ln(1-x)\ \mathrm dx\\
=&4\int_0^1 x^{2k-1}\ln x \ln(1-x^2)\ \mathrm dx
\end{align*} we have that
\begin{align*}
&\int_{0}^{1}\ln(1-x)\ln(1+x) \frac{\ln x\ln(1-x^2)}x \ \mathrm dx \\&=-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\int_{0}^{1}x^{2k-1} \ln x \ln(1-x^2)\ \mathrm dx \\
&=-\frac 1 4\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\left(\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}\right).
\end{align*} The integral can be attacked by considering algebraic identity
$$
ab(a+b) = \frac 1 3 (a+b)^3 - \frac {a^3}3 -\frac{b^3}3
$$ with $a=\ln(1-x)$ and $b=\ln(1+x)$, and extant results.
For the sum, after expanding the summand, the only tricky part is
$$
\sum_{k=1}^\infty\frac{H_{2k}H_k}{k^3},
$$ which can be found here. Then, the sum $\sum_{k=1}^\infty \frac{H_{2k}H_k^{(2)}}{4k^2}$ can be evaluated by solving the equation obtained.
Best Answer
Let $\mathcal{I}$ denote the value of the following logarithmic integral:
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\approx-0.47159.$$
It will be shown that
$$\mathcal{I}=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.$$
Making use of the factorizations $\left(1+x^{3}\right)=\left(1+x\right)\left(1-x+x^{2}\right)$ and $\left(1-x^{4}\right)=\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)$, we can split up the integral into simpler components as
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{1+x}+\frac{2x-1}{1-x+x^{2}}\right]\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{1+x}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\ &=:\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}.\\ \end{align}$$
Let's focus on the last integral first.
Using the derivatives
$$\frac{d}{dx}\ln{\left(1-x+x^{2}\right)}=\frac{2x-1}{1-x+x^{2}},$$
and
$$\frac{d}{dx}\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}=\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)},$$
we show that the following integral vanishes identically:
$$\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}=\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}\bigg{|}_{0}^{1}=0.$$
Then, $\mathcal{I}_{6}$ can be rewritten in the following way:
$$\begin{align} \mathcal{I}_{6} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1+x^{2}\right)}-\ln{\left(1-x+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(\frac{1+x^{2}}{1-x+x^{2}}\right)}\\ &=\int_{1}^{0}\mathrm{d}y\,\frac{\left(-2\right)}{\left(1+y\right)^{2}}\cdot\frac{\left(1+y\right)\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\cdot\frac{\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\left[\frac{2}{\left(1+y\right)}-\frac{6y}{\left(1+3y^{2}\right)}\right]\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{6y}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\left[\ln{\left(2\right)}+\ln{\left(1+y^{2}\right)}-\ln{\left(1+3y^{2}\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{3}{\left(1+3t\right)}\ln{\left(\frac{2+2t}{1+3t}\right)};~~~\small{\left[y^{2}=t\right]}\\ &=2\ln{\left(2\right)}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1+y\right)}+2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{2}\right)}}{\left(1+y\right)}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{3}{\left(1+3u\right)}\ln{\left(1+u\right)};~~~\small{\left[t=\frac{1-u}{1+3u}\right]}\\ &=2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(1+3(t-1)^{2}\right)}}{t};~~~\small{\left[1+y=t\right]}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(3t^{2}-6t+4\right)}}{t}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4u^{2}-4\sqrt{3}\,u+4\right)}}{u};~~~\small{\left[t=\frac{2u}{\sqrt{3}}\right]}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4\right)}}{u}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(u^{2}-\sqrt{3}\,u+1\right)}}{u}\\ &=2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$
The integral $\mathcal{I}_{3}$ can in turn be reduced to
$$\begin{align} \mathcal{I}_{3} &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\ &=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(1+(y-1)^{2}\right)}}{y};~~~\small{\left[1+x=y\right]}\\ &=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(2-2y+y^{2}\right)}}{y}\\ &=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2-2t\sqrt{2}+2t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{2}\right]}\\ &=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2\right)}}{t}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{2}\,t+t^{2}\right)}}{t}\\ &=\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}.\\ \end{align}$$
Next up, $\mathcal{I}_{5}$ can be reduced to a similar expression involving the same undetermined integral in the expression for $\mathcal{I}_{6}$:
$$\begin{align} \mathcal{I}_{5} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1+x};~~~\small{I.B.P.s}\\ &=-\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(3-3y+y^{2}\right)}}{y};~~~\small{\left[x=y-1\right]}\\ &=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3-3t\sqrt{3}+3t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{3}\right]}\\ &=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3\right)}}{t}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(\frac{1-\sqrt{3}\,u+u^{2}}{u^{2}}\right)}}{u};~~~\small{\left[t=u^{-1}\right]}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}-\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}}{u}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\ &=-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$
The remaining three integrals $\mathcal{I}_{1,2,4}$ each have nice exact values in terms of well-known constants: we have
$$\mathcal{I}_{1}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}=-\operatorname{Li}_{2}{\left(\frac12\right)}=\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12},$$
$$\mathcal{I}_{2}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}=\frac12\ln^{2}{\left(2\right)},$$
and
$$\begin{align} \mathcal{I}_{4} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1-x};~~~\small{I.B.P.s}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y+y^{2}\right)}}{y};~~~\small{\left[x=1-y\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{3}\right)}}{y}-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{3t};~~~\small{\left[y^{3}=t\right]}\\ &~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\ &=-\frac23\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{t}\\ &=\frac23\operatorname{Li}_{2}{\left(-1\right)}\\ &=-\frac{\pi^{2}}{18}.\\ \end{align}$$
Then,
$$\begin{align} \mathcal{I} &=\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}\\ &=\left[\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12}\right]+\frac12\ln^{2}{\left(2\right)}+\mathcal{I}_{3}-\frac{\pi^{2}}{18}\\ &~~~~~-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &~~~~~+2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}+3\mathcal{I}_{3}\\ &~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\left[\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}\right]\\ &~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$
At this point, it will be helpful to introduce a two-variable variant of the dilogarithm function:
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
This function can be shown to satisfy the following pair of formulas useful to the problem at hand:
$$\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}=\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)};~~~\small{\theta\in\left[0,\pi\right]},$$
$$\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}=\left(\frac{\pi}{2}-\theta\right)^{2};~~~\small{\theta\in\left[0,\pi\right]}.$$
Then, for $\theta\in\left(0,\frac{\pi}{2}\right)$, we have
$$\begin{align} \int_{\cos{\left(\theta\right)}}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x} &=\int_{0}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\ &~~~~~-\int_{0}^{\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\ &=-2\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}+2\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}\\ &=-2\left(\frac{\pi}{2}-\theta\right)^{2}+2\left[\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}\right]\\ &=\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}-\left(\frac{\pi}{2}-\theta\right)^{2}.\\ \end{align}$$
We can now complete our calculation for $\mathcal{I}$:
$$\begin{align} \mathcal{I} &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\cos{\left(\frac{\pi}{4}\right)}}^{2\cos{\left(\frac{\pi}{4}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{4}\right)}+x^{2}\right)}}{x}\\ &~~~~~-3\int_{\cos{\left(\frac{\pi}{6}\right)}}^{2\cos{\left(\frac{\pi}{6}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{6}\right)}+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{4}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{4}\right)^{2}\right]\\ &~~~~~-3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{6}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{6}\right)^{2}\right]\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+\frac32\operatorname{Li}_{2}{\left(\frac12\right)}-\frac{3\pi^{2}}{16}\\ &~~~~~-\frac32\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{3}\\ &=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+\frac32\left[\frac{\pi^{2}}{12}-\frac12\ln^{2}{\left(2\right)}\right]-\frac{3\pi^{2}}{16}\\ &=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.\blacksquare\\ \end{align}$$