<s>[INST] 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.
 [/INST] 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
$$ I(1)=2\int_{-1}^1\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}da=2\int_{-1}^1\frac{\frac{\pi}{2}}{1+a^2}da=\frac{\pi^2}{2}.$$
</s>