<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] Though there are already 6 wonderful solutions, I want to share mine with you now. Wish that you can enjoy it.
Letting $x=\cos \theta$ yields
$$
I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\cos \theta+\sin \theta)}{\cos \theta}\sin \theta d \theta
$$
and letting $x=\sin \theta$ yields $$
I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta} \cos \theta d \theta
$$
Combining them gives $$
\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta \cos \theta} d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin 2 \theta)}{\sin 2 \theta} d \theta  \\
&\stackrel{2\theta\mapsto\theta}{=} \frac{1}{2} \int_{0}^{\pi} \frac{\ln (1+\sin \theta)}{\sin \theta} d \alpha \\
&=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin \theta)}{\sin \theta} d \theta \quad \textrm{( By symmetry)}\\
&\stackrel{\theta\mapsto \frac{\pi}{2} -\theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta \\
&=\frac{\pi^{2}}{8}
\end{aligned}
$$
Putting $a=0$ in my answer, we get $2 I=\dfrac{\pi^{2}}{8}$ and hence $\boxed{I=\frac{\pi^{2}}{16}}$.
:|D Wish you enjoy my solution!
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