Integral $I =\int_{0}^{1} \frac{\ln(x+\sqrt{1-x^2})}{x}dx$

Question

I tried using trigonometric substitution like

$x= \sin(\theta)$

then applied the substitution

$t=\pi/2 - x$

to get two same integrals and sum them

but now I'm stuck with

$$2I = \int_{0}^\frac{\pi}{2} \frac{\ln(\sin(\theta)+\cos(\theta))}{\sin(\theta)\cos(\theta)}dx$$

any help please?