<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] 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.
</s>