<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] 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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