Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$
Let
$$$$
\begin{align} I(a)&=\int_0^1 2\, \log\frac{\big(x+a\, \sqrt{1-x^2}\big)}{\big(x-a\, \sqrt{1-x^2}\big)} \frac{x\, dx}{1-x^2} \tag 1\\ \therefore \frac{\partial}{\partial a}I(a) &= 2\, \int_0^1 \left(\frac{\sqrt{1-x^2}}{\big(x+a\, \sqrt{1-x^2}\big)}+\frac{\sqrt{1-x^2}}{\big(x-a\, \sqrt{1-x^2}\big)}\right) \frac{x}{1-x^2}\, dx\\ &= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{\tan{(t)}^2-a^2}\, dt \hspace{100pt}\text{(subst. $x=\sin{t}$)}\\ &= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{(\tan{(t)}^2-a^2)(1+\tan{(t)}^2)} \sec{(t)}^2 \, dt\\ &= \int_0^{\infty} \frac{4\, y^2}{(y^2-a^2)(1+y^2)} \, dy \hspace{82pt}\text{(subst. $\tan{t}=y$)}\\ &= \int_0^{\infty} \frac{1}{1+a^2}\left(\frac{2a}{y-a}-\frac{2a}{y+a}+\frac{4}{1+y^2}\right)\, dy\\ &= \frac{1}{1+a^2}\left(a\, \log\left(\frac{y-a}{y+a}\right)^2 +4\arctan{y}\right)\Bigg|_0^\infty\\ &= \frac{2\, \pi}{a^2+1}\\ \implies I(a)&=2\,\pi\arctan{a}+C \tag 2 \end{align}
Putting $a=0$ in $(1)$ and $(2)$, we find that $C=0$
Hence, $$I(a)=2\,\pi\arctan{a}$$
and the required integral $$I(1)=\frac{\pi^2}{2}$$
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$\ds{\int_{0}^{1} \ln\pars{\bracks{x + \root{1 - x^{2}}}^2 \over \bracks{x-\root{1 - x^2}}^{2}}\, {x\,\dd x \over 1 - x^{2}} = {\pi^{2} \over 2}:\ {\large ?}}$
With $\ds{x \equiv \sin\pars{\theta}}$: \begin{align}&\int_{0}^{1} \ln\pars{\bracks{x + \root{1 - x^{2}}}^2 \over \bracks{x-\root{1 - x^2}}^{2}}\, {x\,\dd x \over 1 - x^{2}} = -\int_{0}^{\pi/2} \ln\pars{\bracks{\tan\pars{\theta} - 1 \over \tan\pars{\theta} + 1 }^{2}}\, \tan\pars{\theta}\,\dd\theta \\[5mm]= &\ -\int_{0}^{\pi/2}\ln\pars{\tan^{2}\pars{\theta - {\pi \over 4}}}\, \tan\pars{\theta}\,\dd\theta =-\int_{-\pi/4}^{\pi/4}\ln\pars{\tan^{2}\pars{\theta}}\, \tan\pars{\theta + {\pi \over 4}}\,\dd\theta \\[5mm]= &\ -\int_{0}^{\pi/4}\ln\pars{\tan^{2}\pars{\theta}}\, \bracks{{1 + \tan\pars{\theta} \over 1 - \tan\pars{\theta}} + {1 - \tan\pars{\theta} \over 1 + \tan\pars{\theta}}}\,\dd\theta \\[3mm]&=-2\int_{0}^{\pi/4}\ln\pars{\tan\pars{\theta}}\, {2\sec^{2}\pars{\theta} \over 1 - \tan^{2}\pars{\theta}}\,\dd\theta =-4\ \overbrace{\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t} ^{\ds{-\,{\pi^{2} \over 8}}} = \bbox[15px,#ffe,border:1px dotted navy]{\ds{\pi^{2} \over 2}} \end{align}
The last integral can be evaluated by expanding $\ds{\pars{1 - t^{2}}^{-1}}$ in powers of $\ds{t}$. Namely,
\begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t} =\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}t^{2n}\,\dd t =\left.\sum_{n = 0}^{\infty}\partiald{}{\mu}\int_{0}^{1}t^{2n + \mu}\,\dd t\, \right\vert_{\,\mu = 0} =-\sum_{n = 0}^{\infty}{1 \over \pars{2n + 1}^{2}} \\[3mm]&=-\sum_{n = 1}^{\infty}{1 \over n^{2}} + \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}} =-\,{3 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}} = -\,{3 \over 4}\,{\pi^{2} \over 6}=\color{#c00000}{-\,{\pi^{2} \over 8}} \end{align}
I hope it is not too late. Let $$ I(a)=\int_0^1 \log\frac{\big(x+a\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}, -1\le a\le 1. $$ Clearly $I(-1)=0$ and \begin{eqnarray} I'(a)&=&2\int_0^1 \frac{x}{(x+a\sqrt{1-x^2})\sqrt{1-x^2}}dx\\ &=&2\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt. \end{eqnarray} Define $$ A=\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt, B=\int_0^{\frac{\pi}{2}}\frac{\cos t}{\sin t+a\cos t}dt $$ and then $$ A+aB=\frac{\pi}{2}, B-aA=-\log |a|.$$ Thus $A=\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}$. However $\frac{a\log |a|}{1+a^2}$ is an odd function of $a$ and hence $$ I(1)=2\int_{-1}^1\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}da=2\int_{-1}^1\frac{\frac{\pi}{2}}{1+a^2}da=\frac{\pi^2}{2}.$$