Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ but that didn't help me much .  After this I obtained integrals of the form
$$
\int_0^1 \log \big[\big(x\pm\sqrt{1-x^2}\big)^2\big] \frac{x\, dx}{1-x^2}
$$
which I am not sure how to handle.  Thanks.
 A: 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
$$ I(1)=2\int_{-1}^1\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}da=2\int_{-1}^1\frac{\frac{\pi}{2}}{1+a^2}da=\frac{\pi^2}{2}.$$
A: So here is a solution which follows the logic of both @gar's and @xpaul's but avoids the issue of poles.
Let $I(a)$ be defined by
$$ I(a) = \int_{0}^{1} \frac{2x}{1-x^{2}} \log \left| \frac{x + \sqrt{1-x^{2}}}{x + a\sqrt{1-x^{2}}} \right| \, dx. $$
The value we want to calculate is $I(-1)$. With the substitution $t = x/\sqrt{1-x^{2}}$ (this substitution is essentially the same as in @Felix's answer), it follows that
$$ I(a)
= \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left| \frac{1+t}{a+t} \right| \, dt
= \Re \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt, $$
where the branch of logarithm is chosen to avoid the upper half-plane. Then by DCT,
$$ I(a)
= \lim_{\epsilon \downarrow 0} \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt =: \lim_{\epsilon \downarrow 0}  I_{\epsilon}(a). $$
Now, for $0 < \epsilon \ll 1$, we have
\begin{align*}
I_{\epsilon}'(a)
&= - \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{(1+t^{2})(t+a)} \, dt \\
&= - \frac{2a}{1+a^{2}} \Re \int_{i\epsilon}^{\infty+i\epsilon} \left( \frac{1}{a} \frac{1}{1+t^{2}} + \frac{t}{1+t^{2}} - \frac{1}{t + a} \right) \, dt \\
&= - \frac{2a}{1+a^{2}} \Re \left( \frac{\pi}{2a} - \frac{1}{a}\arctan(i\epsilon) + \log(a+i\epsilon) - \log\sqrt{1-\epsilon^{2}} \right) \\
&= - \frac{2a}{1+a^{2}} \left( \frac{\pi}{2a} + \log\sqrt{a^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right).
\end{align*}
This gives, together with the initial condition $I_{\epsilon}(1) = 0$, that
$$ I_{\epsilon}(a) = \int_{a}^{1} \frac{2s}{1+s^{2}} \left( \frac{\pi}{2s} + \log\sqrt{s^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right) \, ds. $$
Taking $\epsilon \downarrow$, again by DCT, we get
$$ I(a) = \int_{a}^{1} \frac{\pi + 2s\log|s|}{1+s^{2}} \, ds. $$
Plugging $a = -1$ we finally obtain the desired answer.
A: 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}$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$

$\ds{\int_{0}^{1}
\ln\pars{\bracks{x + \root{1 - x^{2}}}^2 \over \bracks{x-\root{1 - x^2}}^{2}}\,
{x\,\dd x \over 1 - x^{2}} = {\pi^{2} \over 2}:\ {\large ?}}$


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}
