Prove $\int_0^1 \frac{\ln x\ +\ \ln(\sqrt x\ +\sqrt {1+x})}{\sqrt {1-x^2}} dx=0$ If a simple way exists, I am looking to show that
$$\boxed{K=\int_0^1 \frac{\ln(x)+\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx=0}$$
say, with  symmetry, a clever change of  variables, or  integrations by parts, without evaluating  integrals separately. It is similar to @Zacky question, of proving $$\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$$
without calculating separately integrals.
If we  take separately
$$I=\int_0^1 \frac{\ln(x)}{\sqrt {1-x^2}}dx,\>\>\>\>J=\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}}dx$$the integrals $I$ and $J$ define the same series to a sign (two series of opposite sums)
$$ I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}},\>\>\>\>\>J=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$$

Explanation By Fourier series $\ln\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)=\sum_{k=0}^\infty\frac{(2k)!}{4^k(2k+1)(k!)^2}\sin((2k+1)x)$ then $J=\int_0^{\frac{\pi}{2}} \ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)dt=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$
$ I=\int_{0}^{1}\frac{\log(t)}{\sqrt{1-t^{2}}}dt$ We know that  $\frac{1}{\sqrt{1-t^{2}}}=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}}t^{2n} $ and $ \int_{0}^{1}\log(t)t^{2n}dt=-\frac{1}{(2n+1)^{2}}$ then $I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}} $

We can write K as
$$K=\int_0^{\dfrac{\pi}{2}} \Big(\ln(\sin t )+\ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)\Big)dt=0 $$
Remark : Wolframalpha can calculate K, but do not know how to calculate $$\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx$$
I believe that Wolframe uses a simple way to see that $K$  is zero . same observation for the Zacky’s integral

I know this is not exactly what it is asked for.
On the path of Zacky,
\begin{align*}
 J&=\int_0^1\frac{\text{arcsinh}\left(\sqrt{x}\right)}{\sqrt{1-x^2}}dx\\
 &\overset{\text{IBP}}=\Big[\arcsin (x)\text{arcsinh}\left(\sqrt{x}\right)\Big]_0^1-\frac{1}{2}\underbrace{\int_0^1 \frac{\arcsin x}{\sqrt{x}\sqrt{1+x}}dx}_{_{z=\sqrt{\frac{1-x}{1+x}}}}\\
 &=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\arcsin\left(\frac{1-z^2}{1+z^2}\right)}{(1+z^2)\sqrt{1-z^2}}dz\\
 &=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\left(\frac{\pi}{2}-2\arctan z\right)}{(1+z^2)\sqrt{1-z^2}}dz\\
 &=\frac{\pi}{2}\left(\ln\left(1+\sqrt{2}\right)-\sqrt{2}\underbrace{\int_0^1 \frac{z}{(1+z^2)\sqrt{1-z^2}}dz}_{=\text{K}}\right)+2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz\\
 K&=\frac{1}{2\sqrt{2}}\left[\ln\left(\frac{\sqrt{2}-\sqrt{1-z^2}}{\sqrt{2}+\sqrt{1-z^2}}\right)\right]_0^1=\frac{1}{\sqrt{2}}\ln\left(1+\sqrt{2}\right)\\\
 J&=2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz
  \end{align*}
Define on $[0,1]$,
\begin{align*}F(a)&=\int_0^1 \frac{z\arctan (az)}{(1+z^2)\sqrt{1-z^2}}dz\\
F^\prime(a)&=\int_0^1 \frac{z^2}{(1+z^2)(1+a^2z^2)\sqrt{1-z^2}}dz\\
&=-\frac{1}{2}\left[\frac{\sqrt{2}\arctan\left(\frac{z\sqrt{2}}{\sqrt{1-z^2}}\right)-\frac{2}{\sqrt{1+a^2}}\arctan\left(\frac{z\sqrt{1+a^2}}{\sqrt{1-z^2}}\right)}{1-a^2}\right]_{z=0}^{z=1}\\
&=\frac{\pi\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{2(1-a^2)\sqrt{2}}
\end{align*}
Since $F(0)=0$ then,
\begin{align*}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz&=F(1)-F(0)\\
&=\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{1-a^2}da\\
&=\frac{\pi}{2\sqrt{2}}\left[\text{arctanh}\left(\frac{\sqrt{2}a}{\sqrt{1+a^2}}\right)-\text{arctanh}(a)\right]_0^1\\
&=\frac{\pi}{2\sqrt{2}}\lim_{a\rightarrow 1}\ln\left(\frac{\sqrt{1+a^2}+a\sqrt{2}}{1+a}\right)\\
&=\frac{\pi\ln 2}{4\sqrt{2}}
\end{align*}
Therefore,
\begin{align*}\boxed{J=\frac{\pi\ln 2}{2}}\end{align*}
