Here is a solution: Let $I$ denote the integral. Then
\begin{align*}
I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\
&= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\
&= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\
&= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\
&= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du.
\end{align*}
Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$:
$$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1. $$
Then it follows that
\begin{align*}
I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\
&= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}.
\end{align*}
Now we consider the exponential of the partial sum:
\begin{align*}
&\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\
&= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\
&= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\
&= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}.
\end{align*}
In view of the definition of Glaisher-Kinkelin constant $A$, we have
$$ \lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}. $$
This, together with the identity $ \log A = \frac{1}{12} - \zeta'(-1)$, yields
$$ I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right) $$
as desired.
Just some considerations for now.
I have proved here that
$$ \sum_{n=1}^{N}H_n^2 = (N+1) H_N^2-(2N+1)H_N+2N \tag{$\color{blue}{1}$} $$
and we have:
$$ \mathcal{L}^{-1}\left(\frac{\log x+\gamma}{x}\right)(s)=-\log(s)\tag{2} $$
$$ \mathcal{L}^{-1}\left(\frac{\left(\log x+\gamma\right)^2}{x}\right)(s)=-\zeta(2)+\log^2(s)\tag{3} $$
hence the partial sums of the given series can be rearranged as follows:
$$\begin{eqnarray*}\sum_{n=1}^{N}\left[H_n^2-\left(\log n+\gamma+\frac{1}{2n}\right)^2\right]&=&(\color{blue}{1})-\frac{H_N^{(2)}}{4}-\sum_{n=1}^{N}\frac{\log n+\gamma}{n}-\sum_{n=1}^{N}n\frac{(\log n+\gamma)^2}{n}\\&=&(\color{blue}{1})-\frac{H_N^{(2)}}{4}+\int_{0}^{+\infty}\frac{\log(s)(1-e^{-Ns})}{e^s-1}\,ds\\&-&\int_{0}^{+\infty}\frac{e^{-N s} \left(e^{(1+N) s}+N-e^s (1+N)\right)\left(\zeta(2)-\log^2 s\right)}{\left(-1+e^s\right)^2}\,ds\end{eqnarray*}$$
If we find a way to distribute $(\color{blue}{1})$ over the last two integrals, in a way ensuring they are convergent integrals as $N\to +\infty$, we are done. At first sight the closed form of the LHS appears to be related with $\zeta(2)$, $\zeta'(0)=-\frac{1}{2}\log(2\pi)$ and
$$ \zeta''(0)=\frac{\gamma^2}{2}-\frac{\pi ^2}{24}-\frac{1}{2}\log^2(2\pi)+\gamma_1$$
with $\gamma_1$ being a Stieltjes constant. An alternative, brute-force way is just to compute the asymptotic expansions of
$$ \sum_{n=1}^{N}\log^2(n),\qquad \sum_{n=1}^{N}\frac{\log(n)}{n} $$
with the sufficient degree of accuracy (I guess that to stop at the $O\left(\frac{1}{N^3}\right)$ term is enough), that is just a tedious exercise about summation by parts.
Best Answer
Let's denote the integral in question as $$I=\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}dx.\tag1$$ Changing the variable $x=y^{1/10}$ and renaming $y$ back to $x$ we get $$I=\int_0^1\frac{x^{2/5}+x^{1/5}-x^{1/10}-1-\ln\sqrt x}{(x-1)\ln x}dx.\tag2$$ Some elementary transformations show that $$I=\mathcal{J}(2/5)+\mathcal{J}(1/5)-\mathcal{J}(1/10),\tag3$$ where we introduced notation $$\mathcal{J}(q)=\int_0^1\frac{x^q-1-q\,\ln x}{(x-1)\ln x}dx.\tag4$$ The integral $\mathcal{J}(q)$ can be evaluated as follows: $$\begin{align}\mathcal{J}(q)=\int_0^1\int_0^q\frac{x^p-1}{x-1}dp\,dx=\int_0^q\underbrace{\int_0^1\frac{x^p-1}{x-1}dx}_{\text{DLMF 5.9.16}}\,dp\\=\int_0^q H_p\,dp=q\cdot\gamma+\ln\Gamma(q+1),\end{align}\tag5$$ where $H_p$ are harmonic numbers: $H_p$$\,=\,$$\gamma$$\,+\,$$\psi_0$$(p+1)$, and $\psi_0$ is the digamma function: $\psi_0(x)=\frac{d}{dx}\ln\,$$\Gamma$$(x)$. Let me mention that the formula DLMF 5.9.16 becomes particularly obvious for positive integer $p$, when $H_p=\sum_{n=1}^pn^{-1}$.
Pluging $(5)$ back into $(3)$, we get $$I=\frac12\gamma+\ln\frac45+\ln\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}.\tag6$$ From the formula $(74)$ on this MathWorld page we know that $$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}=\frac{\sqrt[5]2\,\sqrt\pi}{\sqrt[4]5\,\sqrt\phi}.\tag7$$ (see the paper Raimundas Vidūnas, Expressions for values of the gamma function for a proof).
Making use of this formula, we get the final result $$I=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi.\tag8$$