How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$

How to prove that

$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$

My attempt:

We have for $|x|\leq1$ $$\tanh^{-1}(x)=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$$ and :
\begin{align*} \displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\ \mathrm{d}x&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\mathrm{d}x\\ &=\displaystyle\sum_{n=0}^{\infty}\frac{1}{2n+1}\displaystyle\int_0^{\sqrt{2}-1}x^{2n}\mathrm{d}x\\ &=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\ &=S\\ \end{align*} So :

\begin{align*} S&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\ &=\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\mathrm{d}x\\ &=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{2x}\left(\log(1+x)-\log(1-x)\right)\mathrm{d}x\\ &=\frac{1}{2}(J_1-J_2) \end{align*} Where: \begin{align*} J_1&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1+x)}{x}\mathrm{d}x\\ &=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\displaystyle\int_0^{\sqrt{2}-1}x^n\mathrm{d}x\\ &=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\ \end{align*} And: \begin{align*} J_2&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1-x)}{x}\mathrm{d}x\\ &=\displaystyle\int_0^{\sqrt{2}-1}-\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n+1}dx\\ &=-\displaystyle\sum_{n=0}^{\infty}\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]_0^{\sqrt{2}-1}\\ &=-\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\ \end{align*} Finally we find :

$$S=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}((-1)^n+1)}{(n+1)^2}\right)$$

But I could not find a way to calculate $S$. Any help please? and thank's in advance.

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 $$\frac{\pi^2}{12}+\frac{1}{4}\ln^2(\sqrt 2-1)+\frac{1}{2}\left(-\frac{\pi^2}{24}-\ln^2(\sqrt 2-1)\right)=\frac{\pi^2}{16}-\frac{1}{4}\ln^2\left(\sqrt 2-1\right)$$

$$\color{red}{\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)=\frac{1}{2}\operatorname{Li}_2(z^2)}$$ and for $z=\frac{1}{\sqrt 2}$ we have $$\color{red}{\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{\pi^2}{12}-\ln^2(2)\right)}$$

No need a sledgehammer to crack a nut.

\begin{align} J&=\int_0^{\sqrt{2}-1} \frac{\text{arctanh } x}{x}\,dx\\&=-\frac{1}{2}\int_0^{\sqrt{2}-1}\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}\,dx\\ &\overset{y=\frac{1-x}{1+x}}=-\int_{\sqrt{2}-1}^1 \frac{\ln y}{1-y^2}dy\\ &=\int_0^{\sqrt{2}-1} \frac{\ln y}{1-y^2}dy-\int_0^1 \frac{\ln y}{1-y^2}dy\\ &\overset{\text{IBP}}=\frac{1}{2}\left[\ln\left(\frac{1+y}{1-y}\right)\ln y\right]_0^{\sqrt{2}-1}+\frac{1}{2}\int_0^{\sqrt{2}-1} \frac{\ln\left(\frac{1-y}{1+y}\right)}{y}dy-\int_0^1 \frac{\ln y}{1-y^2}dy\\ &=-\frac{1}{2}\ln^2\left(\sqrt{2}-1\right)-J-\int_0^1 \frac{\ln y}{1-y^2}dy\\ \end{align} And, \begin{align}\int_0^1 \frac{\ln y}{1-y^2}dy&=\int_0^1 \frac{\ln y}{1-y}dy-\int_0^1 \frac{u\ln u}{1-u^2}du\\ &\overset{y=u^2}=\int_0^1 \frac{\ln y}{1-y}dy-\frac{1}{4}\int_0^1 \frac{\ln y}{1-y}dy\\ &=\frac{3}{4}\int_0^1 \frac{\ln y}{1-y}dy\\ &=\frac{3}{4}\times -\frac{\pi^2}{6}\\ &=-\frac{\pi^2}{8} \end{align} Therefore, $\displaystyle \boxed{J=\frac{\pi^2}{16}-\frac{1}{4}\ln^2\left(\sqrt{2}-1\right)}$

NB:

I assume that $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$ and, observe that if $\alpha=\sqrt{2}-1$ then $\displaystyle \dfrac{1-\alpha}{1+\alpha}=\alpha$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} S & \equiv \sum_{n = 0}^{\infty}{\pars{\root{2} - 1}^{2n + 1} \over \pars{2n + 1}^{2}} = \sum_{n = 1}^{\infty}{\pars{\root{2} - 1}^{n} \over n^{2}}\,{1^{n} - \pars{-1}^{n} \over 2} \\[5mm] & = {1 \over 2}\sum_{n = 1}^{\infty}{\pars{\root{2} - 1}^{n} \over n^{2}} - {1 \over 2}\sum_{n = 1}^{\infty}{\pars{1 - \root{2}}^{n} \over n^{2}} \\[5mm] & = {\mrm{Li}_{2}\pars{\root{2} - 1} - \mrm{Li}_{2}\pars{1 - \root{2}} \over 2} \approx 0.4226 \end{align} $\ds{\mrm{Li}_{s}}$ is the Polylogarithm function.

$\color{red}{\tt\large See}$ the nice ${\tt @Naren}$ answer who already worked out the details toward the final answer

$\ds{\bbox[10px,#ffd,border:2px groove navy]{{\pi^{2} \over 16} - {1 \over 4}\,\ln^{2}\pars{\root{2} - 1}}}$