$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I will evaluate the sum at the end of Jim Belk's post.
There is clearly a lot more going on here; I hope a number theorist will come and explain what is really going on.
I will be sloppy about convergence issues.
Set
$$S:= \sum_{k,n=1}^{\infty} \frac{(-1)^{k+n}}{k^2+4kn+n^2}.$$
We set $m = k+2n$, in order to diagonalize the quadratic form:
$$S = \sum_{0 < 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2}.$$
It turns out to be more elegant to work with
$$T : = \sum_{0 \leq 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2} = S + \sum_{m=1}^{\infty} \frac{(-1)^m}{m^2} = S - \frac{\pi^2}{12}.$$
Let $R$ be the ring $\ZZ[\sqrt{3}]$. Let $D \subset R$ be $\{ m+n\sqrt{3} : 0 \leq 2n < m \}$. So, simply as a matter of formal rewriting
$$T = \sum_{m+n \sqrt{3} \in D} \frac{(-1)^{m+n}}{m^2-3 n^2}.$$
We will now try to interpret each of these quantities in terms of the ring $R$. Recall that, for $m+n \sqrt{3} \in R$, the norm $N(m+n \sqrt{3})$ is $m^2 - 3 n^2$.
Define $R_{+} = \{ a \in R : N(a) > 0\}$ and $R_{-} = \{ a \in R : N(a) < 0 \}$.
For $m+n \sqrt{3} \in R$, define $\sigma_2(m+n \sqrt{3}) = (-1)^{m+n}$.
Let $\Gamma$ denote the unit group of $R$. Explicitly, $\Gamma$ is $\{ \pm 1 \} \times (2+\sqrt{3})^{\ZZ}$.
Note that multiplication by $\Gamma$ takes $R_{+}$ and $R_{-}$ to themselves. Here is the first miracle: $D$ is a fundamental domain for the action of $\Gamma$ on $R_{+}$! For example, multiplication by $2+\sqrt{3}$ maps the ray $\RR_{\geq 0} (1+0 \sqrt{3})$ bounding one side of $D$ to the ray $\RR_{\geq 0} (2+\sqrt{3})$ on the other side. Moreover, multiplication by units leaves $N( \ )$ and $\sigma_2(\ )$ unchanged.
So we can view the sum as
$$T = \sum_{a \in R_{+}/\Gamma} \frac{\sigma_2(a)}{N(a)}$$
where the sum means to pick one representative for each orbit of the $\Gamma$ action.
I'd rather sum over $R$ than $R_{+}$. Define $\chi_{\infty}$ to be $\pm 1$ on $R_{\pm}$. So we can rewrite
$$ T = \frac{1}{2} \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \left( 1+\chi_{\infty}(a) \right)}{|N(a)|} = \frac{1}{2}\left( \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a)}{|N(a)|} + \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \chi_{\infty}(a)}{|N(a)|} \right) = : \frac{1}{2} (U+V).$$
Now, $a$ and $b$ in $R$ are in the same $\Gamma$ orbit if and only if the ideals $(a)$ and $(b)$ are equal. So the above sums are running over all principal ideals of $R$.
Moreover, $R$ is a PID! And, finally, for a principal ideal $I = (a)$, we have $N(I) = |N(a)|$. So we can write:
$$U = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)} \quad V = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)} .$$
We set
$$U(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)^s} \quad V(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)^s} .$$
Note that $\sigma_2(a)$ is $1$ if and only if $1+\sqrt{3}$ divides $a$.
So we have the Euler factorization
$$U(s) = \left(-1 + 2^{-s} + 2^{-2s} + 2^{-3s} + \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-N(\pi)^{-s}} =$$
$$\frac{-1+2^{-s}+2^{-2s}+ 2^{-3s} + \cdots}{1+2^{-s}+2^{-2s}+2^{-3s}+\cdots} \prod_{\pi} \frac{1}{1-N(\pi)^{-s}} =(-1+2\cdot 2^{-s}) \zeta_R(s).$$
where $\pi$ runs over prime ideals of $R$.
So
$$\lim_{s \to 1^{+}} U(s) = \lim_{s \to 1^{+}} \frac{-1+2 \cdot 2^{-s}}{s-1} \lim_{s \to 1^{+}} (s-1) \zeta_R(s) = - \log 2 \lim_{s \to 1^{+}} (s-1) \zeta_R(s).$$
From the class number formula, this last limit is
$$\frac{2^2 \log (2+\sqrt{3})}{2 \cdot \sqrt{12}}.$$
So
$$U(1) = - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}}.$$
We now run the same trick with $V$. Again, we start with the Euler product:
$$V(s) = \left(-1 - 2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} - \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} =$$
$$\frac{-1 - 2^{-s} + 2^{-2s} - 2^{-3s} +2^{-4s} - \cdots}{1-2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} - \cdots} \prod_{\pi} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} =
\left( -1 - 2^{1-s} \right) L(s, \chi_{\infty}).$$
So
$$V = V(1) = - 2 L(1, \chi_{\infty}).$$
We now must evaluate $L(1, \chi_{\infty})$. This $L$-function is defined by a sum over the ideals of $R$; we will rewrite it in terms of $L$-functions for $\ZZ$.
Let $p \geq 5$ be prime. I claim that $p$ splits in $R$ if and only if $p \equiv \pm 1 \bmod 12$. For such a $p$, if we write $p = \pi \bar{\pi}$, I claim that $N(\pi) = N(\bar{\pi}) = p$ if $p \equiv 1 \bmod 12$ and $N(\pi) = N(\bar{\pi}) = -p$ if $p \equiv -1 \bmod 12$. Proof Sketch: The prime $p$ splits in $R$ if and only if $\left( \frac{3}{p} \right) = 1$, which is easily computed to be equivalent to $p \equiv \pm 1 \bmod 12$. Since $R$ is a PID, we can write such a $p$ as $\pi \bar{\pi}$ for $\pi$ and $\bar{\pi} \in R$. Then $N(\pi) = N(\bar{\pi}) = \pm p$. But, for $a=m+n \sqrt{3} \in R$, we have $N(a) = m^2-3 n^2 \not \equiv -1 \mod 3$. So only one of the two possibilities for $\pm p$ can occur. $\square$
So $L(1, \chi_{\infty})$ is
$$\left(1+2^{-1} \right)^{-1} \left( 1+3^{-1} \right)^{-1} \prod_{p \equiv 1 \bmod 12} \left( 1- p^{-1} \right)^{-2} \prod_{p \equiv -1 \bmod 12} \left( 1+ p^{-1} \right)^{-2} \prod_{p \equiv \pm 5 \bmod 12} \left( 1- p^{-2} \right)^{-1}.$$
Define $\chi_4(n)$ to be $0$ if $n$ is even, $1$ if $n \equiv 1 \bmod 4$ and $-1$ is $n \equiv -1 \bmod 4$. Define $\chi_3(n)$ to be $1$, $-1$ or $0$ according to whether $n \equiv 1$, $2$ or $0 \bmod 3$. Then we have
$$L(1, \chi_{\infty}) = \prod_p \left( 1- \chi_4(p) p^{-1} \right)^{-1} \prod_p \left( 1-\chi_3(p) p^{-1} \right)^{-1} = \left( \sum_{n=1}^{\infty} \frac{\chi_4(n)}{n} \right) \left( \sum_{n=1}^{\infty} \frac{\chi_3(n)}{n} \right) .$$
The sum $\sum \chi_4(n) = 1-1/3+1/5-1/7 + \cdots$ is well known to be $\pi/4$. The sum $\sum \chi_3(n)/n$ is only slightly less well known; it is $\pi/(3 \sqrt{3})$.
So I get $L(1,\chi_{\infty}) = \frac{\pi}{4} \frac{\pi}{3 \sqrt{3}} = \frac{\pi^2}{12 \sqrt{3}}$ and $V = - \frac{\pi^2}{6 \sqrt{3}}$.
Putting it all together,
$$T = \frac{1}{2} \left( - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}} - \frac{\pi^2}{6 \sqrt{3}} \right) = - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} - \frac{\pi^2}{12 \sqrt{3}} $$
$$S = \frac{\pi^2}{12} - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} - \frac{\pi^2}{12 \sqrt{3}} .$$
The original integral is
$$\frac{1}{2} \left( \log(2)^2 -2 \sqrt{3} S \right) = \frac{\log(2)^2}{2} - \frac{\pi^2 \sqrt{3}}{12} + \frac{\log 2 \log(2+\sqrt{3})}{2} + \frac{\pi^2}{12}$$
$$=\frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \frac{\log(4 + 2 \sqrt{3})}{2} = \frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \log(1+\sqrt{3})$$
as desired.
Let $\mathcal{S}$ denote the sum of the following infinite series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}},$$
where here $\overline{H}_{n}$ denotes the $n$-th skew-harmonic number, defined for each positive integer $n$ by the finite sum
$$\overline{H}_{n}:=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k};~~~\small{n\in\mathbb{N}}.$$
As you point out, it can be shown that the skew-harmonic numbers have the following generating function
$$\sum_{n=1}^{\infty}z^{n}\,\overline{H}_{n}=\frac{\ln{\left(1+z\right)}}{1-z};~~~\small{|z|<1}.$$
Using this, as well as the fact that
$$n\binom{2n}{n}\operatorname{B}{\left(n,n\right)}=2;~~~\small{n\in\mathbb{N}},$$
we find
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\
&=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\operatorname{B}{\left(n,n\right)}\\
&=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\int_{0}^{1}\mathrm{d}t\,t^{n-1}\left(1-t\right)^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}t^{n-1}\left(1-t\right)^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left[-t\left(1-t\right)\right]}\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{(-1)}{t\left(1-t\right)}\cdot\frac{\ln{\left(1-t(1-t)\right)}}{1+t(1-t)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)\left(1+t-t^{2}\right)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1-t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1-u+u^{2}\right)}}{u};~~~\small{\left[t=1-u\right]}\\
&~~~~~+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}.\\
\end{align}$$
Consider the following two-variable variant of the dilogarithm:
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\theta\right)}+t^{2}\right)}}{t};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
It can be shown that
$$\operatorname{Li}_{2}{\left(1,\theta\right)}=\frac14\left(\pi-\theta\right)^{2}-\frac{\pi^{2}}{12};~~~\small{0<\theta<\pi}.$$
The following particular case can be used to evaluate the first integral of $\mathcal{S}$:
$$-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}=2\operatorname{Li}_{2}{\left(1,\frac{\pi}{3}\right)}=\frac12\left(\pi-\frac{\pi}{3}\right)^{2}-\frac{\pi^{2}}{6}=\frac{\pi^{2}}{18}.$$
Let $\mathcal{I}$ denote the value of the remaining definite integral:
$$\mathcal{I}:=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\approx0.0776351.$$
We find that $\mathcal{I}$ can be expressed in terms of the two-variable dilogarithm as follows:
$$\begin{align}
\mathcal{I}
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\frac12\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}-\frac12\int_{\frac12}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}};~~~\small{symmetry}\\
&=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(\frac34+\left(\frac12-t\right)^{2}\right)}}{\frac54-\left(\frac12-t\right)^{2}}\\
&=-\int_{0}^{\frac12}\mathrm{d}u\,\frac{\ln{\left(\frac34+u^{2}\right)}}{\frac54-u^{2}};~~~\small{\left[t=\frac12-u\right]}\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\sqrt{5}}}\mathrm{d}v\,\frac{\ln{\left(\frac34+\frac54v^{2}\right)}}{1-v^{2}};~~~\small{\left[u=\frac{\sqrt{5}}{2}v\right]}\\
&=-\frac{1}{\sqrt{5}}\int_{1}^{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}\mathrm{d}w\,\frac{(-2)}{(1+w)^{2}}\cdot\frac{(1+w)^{2}}{2w}\ln{\left(\frac{3+5\left(\frac{1-w}{1+w}\right)^{2}}{4}\right)};~~~\small{\left[v=\frac{1-w}{1+w}\right]}\\
&=-\frac{1}{\sqrt{5}}\int_{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{3(1+w)^{2}+5(1-w)^{2}}{4(1+w)^{2}}\right)}\\
&=-\frac{1}{\sqrt{5}}\int_{\frac{\sqrt{5}-1}{\sqrt{5}+1}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)}\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)};~~~\small{\phi:=\frac{1+\sqrt{5}}{2}}\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\left[\ln{\left(2\right)}-2\ln{\left(1+w\right)}+\ln{\left(1-\frac12w+w^{2}\right)}\right]\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(2\right)}}{w}+\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{2\ln{\left(1+w\right)}}{w}\\
&~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-\frac12w+w^{2}\right)}}{w}\\
&=-\frac{\ln{\left(2\right)}}{\sqrt{5}}\ln{\left(\phi^{2}\right)}-\frac{2}{\sqrt{5}}\int_{-1}^{-\phi^{-2}}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x};~~~\small{\left[w=-x\right]}\\
&~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-2w\cos{\left(\theta\right)}+w^{2}\right)}}{w};~~~\small{\left[\theta:=\frac{\pi}{2}-\arcsin{\left(\frac14\right)}\right]}\\
&=-\frac{2\ln{\left(2\right)}\ln{\left(\phi\right)}}{\sqrt{5}}+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(-1\right)}\right]\\
&~~~~~+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(1,\theta\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\
&=\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right].\\
\end{align}$$
Hence,
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=\frac{\pi^{2}}{18}-\mathcal{I}\\
&=\frac{\pi^{2}}{18}-\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\
&=\frac{\pi^{2}}{18}+\frac{2}{\sqrt{5}}\ln{\left(2\right)}\ln{\left(\phi\right)}-\frac{1}{2\sqrt{5}}\left(\pi-\theta\right)^{2}-\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}+\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}.\\
\end{align}$$
Best Answer
I shall be continuing from the last two integrals of your work $$\int_0^{\sqrt 2-1} \frac{\tanh^{-1}x}{x} dx =\frac{1}{2}\int_0^{\sqrt 2-1} \left(\frac{\log(1+x)}{x}-\frac{\log(1-x)}{x}\right)dx$$ These two integrals posses non-elementary antiderivatives in terms of special function Dilogarithm function so by definition we have $$=\frac{1}{2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\bigg|_0^{\sqrt 2-1}\right)=\frac{1}{2}\left(\operatorname{Li}_2\left(\sqrt2 -1\right)-\operatorname{Li}_2\left(1-\sqrt 2\right)\right)\cdots(1)$$ Using the last identity [3] we have $$\operatorname{Li}_2(1-\sqrt 2)= -\frac{\pi^2}{6}-\frac{1}{2}\ln^2(\sqrt 2-1)+\operatorname{Li}_2\left(\frac{1}{1-\sqrt 2}\right)=\frac{\pi^2}{6}-\frac{1}{2}\ln^2(\sqrt 2-1)-\operatorname{Li}_2(-1-\sqrt 2)$$ plugging the obtained to $(1)$ we have then $$\frac{\pi^2}{12}+\frac{1}{4}\ln^2(\sqrt 2-1)+\frac{1}{2}\left(\operatorname{Li}_2(\sqrt 2-1)+\operatorname{Li}_2(-\sqrt 2-1)\right)\cdots(2)$$ let $u= \sqrt 2-1 $ and $v=-1-\sqrt 2$ also $uv= -(\sqrt 2-1)(\sqrt 2+1)=-1$.
Using the Abel identity $$\operatorname{Li}_2(u)+\operatorname{Li}_2(v)=\operatorname{Li}_2(uv)+\operatorname{Li}_2\left(\frac{u-uv}{1-uv}\right)+ \operatorname{Li}_2\left(\frac{v-uv}{1-uv}\right)+\ln\left(\frac{1-u}{1-uv}\right)\ln\left(\frac{1-v}{1-uv}\right)$$ plugging the assigned values of $u$ and $v$ we have $$\operatorname {Li}_(\sqrt 2-1)+\operatorname{Li_2}(-1-\sqrt 2)=-\operatorname{Li}_2(-1)+\color{red}{\operatorname{Li}_2\left(\frac{1}{\sqrt 2}\right)+\operatorname{Li}_2\left(-\frac{1}{\sqrt 2}\right)}+\ln\left(\frac{\sqrt 2-1}{\sqrt{2}}\right)\ln\left(\frac{\sqrt{2}+1}{\sqrt 2} \right) =-\frac{\pi^2}{12}+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)-\ln^2(\sqrt {2}-1)+\frac{1}{2}\ln^2(2)=-\frac{\pi^2}{12}+\color{red}{\frac{\pi^2}{24}-\frac{1}{2}\ln^2(2)}-\ln^2(\sqrt 2-1)+\frac{1}{2}\ln^2(2)=-\frac{\pi^2}{24}-\ln^2(\sqrt 2-1)\cdots(2)$$ plugging back to $(2)$ we have $$\frac{\pi^2}{12}+\frac{1}{4}\ln^2(\sqrt 2-1)+\frac{1}{2}\left(-\frac{\pi^2}{24}-\ln^2(\sqrt 2-1)\right)=\frac{\pi^2}{16}-\frac{1}{4}\ln^2\left(\sqrt 2-1\right)$$
We use identity
$$\color{red}{\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)=\frac{1}{2}\operatorname{Li}_2(z^2)}$$ and for $z=\frac{1}{\sqrt 2}$ we have $$\color{red}{\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{\pi^2}{12}-\ln^2(2)\right)}$$