<s>[INST] How would I integrate $\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$? $$\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$$
I have known this equals to $\frac{\pi^2}{16}$, but I cannot prove it.
 [/INST] Note
\begin{align}
I=&\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx\\
= &\int_0^1 \frac{\frac12\ln(1-x^2)+\ln(1+ \frac x{\sqrt{1-x^2}})}{x}\,dx=\frac12I_1+I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln(1-x^2)}{x}\,dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&= \int_0^1 \frac{\ln(1-x)}{x}\,dx+ \int_0^1 \frac{\ln(1+x)}{x}\,dx
= -\int_0^1 \frac{\ln(1+x)}{x}\,dx\\
I_2 &= \int_0^1 \frac{\ln \left(1+\frac x{\sqrt{1-x^2}}\right)}x
\overset{x=\frac t{\sqrt{1+t^2}}}=\int_0^1\frac{\ln(1+t)}{t(1+t^2)}dt+ \int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t^2)}dt}\\
&=\int_0^1 \frac{\ln(1+t)}t dt - \frac14 \int_0^1 \frac{\ln t}{1+t}dt
=\frac54 \int_0^1 \frac{\ln(1+t)}t dt 
\end{align}
Substitute $I_1$ and $I_2$ into (1) to obtain
$$I= \frac34 \int_0^1 \frac{\ln(1+t)}t dt =\frac34\cdot \frac{\pi^2}{12}=\frac{\pi^2}{16}
$$
$\int_0^1 \frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$
</s>