<s>[INST] Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
 [/INST] Let $I(a)=\int_0^1 \frac{\ln(a x+\sqrt{1-x^2})}{x}dx$. Then, $I(0) = \int_0^1 \frac{\ln\sqrt{1-x^2}}{x}dx
\overset{x^2\to x}
=-\frac{\pi^2}{24}$
$$I’(a)=\int_0^1 \frac{dx}{a x+\sqrt{1-x^2}}
=\frac1{1+a^2}\left(\frac\pi2+a\ln a \right)$$
and
$$\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx
=I(1)= I(0)+\int_0^1I’(a)da \\
\hspace{20mm}= -\frac{\pi^2}{24}+ \frac\pi2\int_0^1 \frac{da}{1+a^2}
+ \int_0^1 \frac{a\ln a}{1+a^2}da= \frac{\pi^2}{16}
$$
</s>