Proving $\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$ 
$$\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$$

At first, I think it can be calculated like the following one with differential method.

$$\int_0^1\frac{x-\log\left({x+\sqrt{1-x^2}}\right)}{x}dx$$ $\\$

But, I'm wrong. I try my best to do it with differential method, but failed. Who can help me to prove it? Thank you.
 A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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}
A: This integral can be evaluated just by a few substitutions. Letting $x=\sin t$, $u=\tan\frac{t}{2}$ and $w=u^2$, we have
\begin{eqnarray}
&&\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx\\
&=&\int_0^{\frac{\pi}{2}}\frac{\ln 2-\ln(1+\cos t)}{\sin t}\cos t dt\\
&=&\int_0^1\frac{\ln 2-\ln(1+\frac{1-u^2}{1+u^2})}{\frac{2u}{1+u^2}}\frac{1-u^2}{1+u^2} \frac{2}{1+u^2}du\\
&=&\int_0^1\frac{(1-u^2)\ln(1+u^2)}{u(1+u^2)}du\\
&=&\frac12\int_0^1\frac{(1-w)\ln(1+w)}{w(1+w)}dw\\
&=&\frac12\int_0^1\frac{1}{w}\ln(1+w)dw-\int_0^1\frac{\ln(1+w)}{1+w}dw\\
&=&\frac{\pi^2}{24}-\frac{1}{2}\ln^22.
\end{eqnarray}
A: Let $(1+\sqrt{1-x^2})/2=t$, we get
$$\int_0^1\frac{\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}dx=\frac12\int_{1/2}^1\frac{\ln(t)}{1-t}dt-\frac12\int_{1/2}^1\frac{\ln(t)}{t}dt=-\frac14\zeta(2)+\frac12\ln^2(2).$$
A different way is by setting $\sqrt{1-x^2}=y$ then subbing $y=(1-t)/(1+t)$:
$$\int_0^1\frac{\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}dx=\int_0^1\frac{y\ln\left(\frac{1+y}{2}\right)}{1-y^2}dy$$
$$=\int_0^1\frac{\ln(1+t)}{1+t}dt-\frac12\int_0^1\frac{\ln(1+t)}{t}dt$$
$$=\frac12\ln^2(2)-\frac14\zeta(2).$$
A: 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
$$I=\int_0^1\frac{\ln 2-\ln\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\ln^22\right)}{24}$$
