Proving $\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$
Denote the integral as $I$ and make the substitution $\sqrt{1-x^2}\mapsto x$, we get $$ \begin{align} I&=-\int_0^1\frac{x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\,dx\\ &=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1+x}\,dx-\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}\,dx\\ &=I_1-I_2 \end{align} $$ Evaluation of $I_1$ $$ \begin{align} I_1 &=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1+x}\,dx\\ &=\frac{1}{2}\int_0^1\frac{\ln\left(1+x\right)}{1+x}\,dx-\frac{1}{2}\int_0^1\frac{\ln2}{1+x}\,dx\\ &=-\frac{\ln^22}{4} \end{align} $$ Evaluation of $I_2$ $$ \begin{align} I_2 &=\frac{1}{2}\int_{1/2}^1\frac{\ln x}{1-x}\,dx\qquad\Rightarrow\qquad\frac{1+x}{2}\mapsto x\\ &=\frac{1}{2}\int_{1/2}^1\sum_{n=1}^\infty x^{n-1}\ln x\,dx\\ &=\frac{1}{2}\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln x\,dx\\ &=\frac{1}{2}\sum_{n=1}^\infty\left[\frac{\ln2}{n\,2^n}+\frac{1}{n^2\,2^n}-\frac{1}{n^2}\right]\\ &=\frac{1}{2}\left[\ln^22+\text{Li}_2\left(\frac{1}{2}\right)-\zeta(2)\right]\\ &=\frac{\ln^22}{4}-\frac{\pi^2}{24} \end{align} $$ Thus $$I=\int_0^1\frac{\ln 2-\ln\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\ln^22\right)}{24}$$
First let $x=\sqrt{1-y^2}$ and the integral becomes
$$\int_0^1 dy \frac{y}{1-y^2} \left [\log{2} - \log{(1+y)} \right ] $$
Then sub $y=1-2 u$, use partial fractions, and then use $u \mapsto 1-u$ to get
$$-\frac12 \int_{1/2}^1 du \frac{\log{u}}{u} - \frac12 \int_{1/2}^1 du \frac{\log{u}}{1-u} $$
which is equal to
$$-\frac14 \log^2{2} +\frac{\pi^2}{24} - \frac14 \log^2{2} $$
The result follows.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + \root{1 - x^{2}}} \over x}\,\dd x ={\pi^{2} - 12\ln^{2}\pars{2} \over 24}:\ {\large ?}}$.
\begin{align}&\color{#66f}{\large% \int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + \root{1 - x^{2}}} \over x}\,\dd x} =-\ \overbrace{\int_{0}^{1}\ln\pars{1 + \root{1 - x^{2}} \over 2}\,{\dd x \over x}} ^{\dsc{x} \equiv \dsc{\sin\pars{\theta}}} \\[5mm]&=-\ \overbrace{\int_{0}^{\pi/2}\ln\pars{1 + \cos\pars{\theta} \over 2}\, {\cos\pars{\theta}\,\dd\theta \over \sin\pars{\theta}}} ^{\dsc{t}\equiv\dsc{\tan\pars{\theta \over 2}}}\ =\ \overbrace{% \int_{0}^{1}\ln\pars{1 + t^{2}}\,{1 - t^{2} \over \pars{1 + t^{2}}t}\,\dd t} ^{\dsc{t} \mapsto \dsc{t^{1/2}}} \\[5mm]&=\half\int_{0}^{1}\ln\pars{1 + t}\,\ \overbrace{{1 - t \over \pars{1 + t}t}}^{\dsc{{1 \over t} - {2 \over 1 + t}}} \,\dd t =\half\int_{0}^{1}{\ln\pars{1 + t} \over t}\,\dd t -\int_{0}^{1}{\ln\pars{1 + t} \over 1 + t}\,\dd t \\[5mm]&=\half\int_{0}^{-1}{\ln\pars{1 - t} \over t}\,\dd t -\bracks{\half\,\ln^{2}\pars{1 + t}}_{0}^{1} =-\,\half\int_{0}^{-1}\Li{2}'\pars{t}\,\dd t -\half\,\ln^{2}\pars{2} \\[5mm]&=-\,\half\,\ \overbrace{\Li{2}\pars{-1}} ^{\dsc{-\,{\pi^{2} \over 12}}}\ -\ \half\,\ln^{2}\pars{2} =\color{#66f}{\large{\pi^{2} - 12\ln^{2}\pars{2} \over 24}} \end{align}