Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.
We can break up the integral into
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x-\int^1_0\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x\\
=&\frac{15}{16}\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{16}\int^1_0\frac{x^{-1/2}\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\
=&\frac{15}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(1,0^{+})-\frac{1}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(0.5,0^{+})
\end{align}
After differentiating and expanding at $b=0$ (with the help of Mathematica),
\begin{align}
&\frac{\partial^4\beta}{\partial a^3 \partial b}(a,0^{+})\\
=&\left[\frac{\Gamma(a)}{\Gamma(a+b)}\left(\frac{1}{b}+\mathcal{O}(1)\right)\left(\left(-\frac{\psi_4(a)}{2}+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)\right)b+\mathcal{O}(b^2)\right)\right]_{b=0}\\
=&-\frac{1}{2}\psi_4(a)+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)
\end{align}
Therefore,
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&-\frac{15}{32}\psi_4(1)+\frac{45}{16}\psi_1(1)\psi_2(1)+\frac{1}{32}\psi_4(0.5)+\frac{1}{8}\psi_3(0.5)\ln{2}-\frac{3}{16}\psi_1(0.5)\psi_2(0.5)\\
=&-12\zeta(5)+\frac{3\pi^2}{8}\zeta(3)+\frac{\pi^4}{8}\ln{2}
\end{align}
The relation between $\psi_{m}(1)$, $\psi_m(0.5)$ and $\zeta(m+1)$ is established easily using the series representation of the polygamma function.
We write
$$ S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} $$
for the sum to be computed.
1st Solution. We have
\begin{align*}
S
= \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \left( - k^2 \log \left( 1 - \frac{1}{k^2} \right) - 1 \right).
\end{align*}
In order to compute this, we write $S_K$ for the partial sums of the last step. Then
\begin{align*}
S_K
&= -K + 1 + \sum_{k=2}^{K} k^2 \log \left( \frac{k^2}{(k+1)(k-1)} \right) \\
&= -K + 1 + \sum_{k=2}^{K} 2 k^2 \log k - \sum_{k=3}^{K+1} (k-1)^2 \log k - \sum_{k=1}^{K-1} (k+1)^2 \log k \\
&= -K + 1 + \log 2 - K^2 \log(K+1) + (K+1)^2 \log K \\
&\quad + \sum_{k=2}^{K} (2 k^2 - (k-1)^2 - (k+1)^2 ) \log k \\
&= -K + 1 + \log 2 - K^2 \log\left(1 + K^{-1}\right) + (2K+1)\log K - 2 \log (K!).
\end{align*}
Now by the Stirling's approximation and the Taylor series of $\log(1+x)$,
$$ 2\log (K!) = \left(2K + 1\right) \log K - 2 K + \log(2\pi) + \mathcal{O}(K^{-1}) $$
and
$$ K^2 \log\left(1 + K^{-1}\right) = K - \frac{1}{2} + \mathcal{O}(K^{-1}) $$
as $K \to \infty$. Plugging this back to $S_K$, we get
$$ S_K = \frac{3}{2} - \log \pi + \mathcal{O}(K^{-1}) $$
and the desired identity follows by letting $K\to\infty$.
2nd Solution. We begin by noting that the Taylor expansion of the digamma function
\begin{align*}
\psi(1+z)
&= -\gamma + \sum_{k=1}^{\infty} (-1)^{k-1} \zeta(k+1) z^{k} \\
&= -\gamma + \zeta(2) z - \zeta(3) z^2 + \zeta(4) z^3 - \dots,
\end{align*}
holds for $|z| < 1$. Then by the Abel's Theorem,
\begin{align*}
S
&= \int_{0}^{1} \sum_{n=1}^{\infty} 2 (\zeta(2n)-1) x^{2n+1} \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(1-x) - \frac{2x}{1-x^2} \right) \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(2-x) + \frac{1}{1+x} \right) \, \mathrm{d}x, \tag{1}
\end{align*}
where the identity
$$ \psi(1+z) = \psi(z) + \frac{1}{z} \tag{2} $$
is used in the last step. Then by using the substitution $x\mapsto 1-x$, we get
$$ \int_{0}^{1} x^2 \psi(2-x) \, \mathrm{d}x
= \int_{0}^{1} (1-x)^2 \psi(1+x) \, \mathrm{d}x. $$
Plugging this back to $\text{(1)}$ and performing integration by parts,
\begin{align*}
S
&= \int_{0}^{1} (2x-1) \psi(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x \\
&= -2 \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x.
\end{align*}
Now the integrals in the last step can be computed as
$$ \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x = -1 + \frac{1}{2}\log(2\pi)
\qquad \text{and} \qquad
\int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x = -\frac{1}{2} + \log 2. $$
For instance, the first integral can be computed by writing $\log\Gamma(x+1) = \log\Gamma(x) + \log x$ and applying the Euler's reflection formula. For a more detail, check this posting. Finally, plugging these back to $S$ proves the desired identity.
Best Answer
Not an answer:
$$\sum _{n=2}^{\infty } \frac{1}{n \sqrt{n^2-1}}=\\\int_0^{\infty } \left(\mathcal{L}_n^{-1}\left[\frac{1}{n \sqrt{n^2-1}}\right](x)\right) \sum _{s=2}^{\infty } \exp (-s x) \, dx=\\\int_0^{\infty } \frac{\pi x (I_0(x) \pmb{L}_{-1}(x)-I_1(x) \pmb{L}_0(x)) e^{-x}}{2 \left(-1+e^x\right)} \, dx\approx0.69422401996922706=\sum _{m=0}^{\infty } \frac{(-1)^m \sqrt{\pi } (-1+\zeta (2+2 m))}{\Gamma \left(\frac{1}{2}-m\right) \Gamma (1+m)}$$
Where: $I_0(x)$ is modified Bessel function of the first kind, $\pmb{L}_0(x)$ is modified Struve function.
$\mathcal{L}_n^{-1}$ is Inverse Laplace Transform.
UPDATE:
Using:$$\int_0^1 \frac{1}{\left(n^2-x\right) \sqrt{x (1-x)} \pi } \, dx=\frac{1}{n \sqrt{n^2-1}}$$ we have: $$\int_0^1 \left(\sum _{n=2}^{\infty } \frac{1}{\left(n^2-x\right) \sqrt{x (1-x)} \pi }\right) \, dx=\int_0^1 -\frac{-1+3 x-\pi (-1+x) \sqrt{x} \cot \left(\pi \sqrt{x}\right)}{2 \pi (-((-1+x) x))^{3/2}} \, dx$$
Using:
$$\int_0^{\pi } \frac{2 x}{\pi ^2 \left(n^2-\cos ^2(x)\right)} \, dx=\frac{1}{n \sqrt{n^2-1}}$$
we have:
$$\int_0^{\pi } \left(\sum _{n=2}^{\infty } \frac{2 x}{\pi ^2 \left(n^2-\cos ^2(x)\right)}\right) \, dx=\\\int_0^{\pi } \left(-\frac{3 x \csc ^2(x)}{2 \pi ^2}-\frac{x \cot (\pi \cos (x)) \sec (x)}{\pi }+\frac{3 x \sec ^2(x)}{2 \pi ^2}-\frac{x \csc ^2(x) \sec ^2(x)}{2 \pi ^2}\right) \, dx$$