Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ 
Prove that
  $$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$

I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} \ln(\sin{x})dx=\int_{0}^{\frac{\pi}2} \ln(\cos{x})dx=-\frac{\pi}2\ln2$$
but it didnt work. Any help? 

$$I=\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin(2x)})dx =\frac12 \int_0^\frac{\pi}{2} x'\ln(\sin x+\cos x+\sqrt{\sin (2x)})dx$$
$$\overset{IBP}=\frac12 \int_0^\frac{\pi}{2}x\,\frac{\sin x-\cos x}{\sqrt{\sin(2x)}}dx\overset{x=\arctan t}=\frac{1}{2\sqrt 2}\int_0^\infty \frac{\arctan t}{1+t^2}\frac{t-1}{\sqrt t}dt$$

$$I(a)=\int_0^\infty \frac{\arctan(at)}{1+t^2}\frac{t-1}{\sqrt t}dt\Rightarrow I'(a)=\int_0^\infty \frac{(t-1)\sqrt t}{(1+a^2 t^2)(1+t^2)}dt$$
$$\overset{t=x^2}=\frac{2}{1-a^2}\int_0^\infty \frac{1+a^2 x^2}{1+a^2 x^4}dx-\frac{2}{1-a^2}\int_0^\infty \frac{1+x^2}{1+x^4}dx=\frac{\pi}{\sqrt 2}\frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}$$

$$\Rightarrow I=\frac{\pi}{4}\int_0^1 \frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}da\overset{\sqrt a=x}=\frac{\pi}{2}\int_0^1 \frac{1-x}{(1+x^2)(1+x)}dx=\frac{\pi}{4}\ln 2 $$
