Evaluate $$\int_0^1 \frac{\ln{\left(x+\sqrt{1-x^2}\right)}}{x} \; \mathrm{d}x$$ I tried $x=\sin{t}$ the symmetry with $t=\frac{\pi}{2}-t$ to get $$\int_0^{\frac{\pi}{2}} \frac{\ln{\left(\sin{t}+\cos{t}\right)}}{2\sin{t}\cos{t}} \; \mathrm{d}x=\int_0^{\frac{\pi}{2}} \frac{\ln{\left(\sin{t}+\cos{t}\right)}}{\sin{2t}} \; \mathrm{d}x$$
This looks much nicer I think but now what...? I considered $\sin{t}+\cos{t}=\sqrt{2} \sin{\left(x+\frac{\pi}{4}\right)}$ but that doesnt help. Any hints or suggestions please.
Integral equals about 0.61685028