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?

As James Arathoon mentioned in the comments, by the substitution $t= \sqrt[]{\frac {1-x^2}{x^2}}$ the integral is equal to:
\begin{align}
I:=\int^1_0 \frac{\log(x+\sqrt[]{1-x^2} )}{x}\,dx=\int^\infty_0 \frac{t\log\left( \frac{t+1}{\sqrt[]{t^2+1}}\right)}{t^2+1}\,dt 
\end{align}
One can rewrite it a bit:
\begin{align}
I=\frac 1 2 \int^\infty_0 \frac{t\log\left( \frac{(t+1)^2}{t^2+1}\right)}{t^2+1}\,dt = \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2t}{t^2+1}\right)}{t^2+1}\,dt
\end{align}
Now define the following function $F:[0, 1]\to\mathbb R$ as follows:
\begin{align}
F(a) :=  \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2at}{t^2+1}\right)}{t^2+1}\,dt
\end{align}
Using Feynman's Trick one gets:
\begin{align}
F'(a) = \int^\infty_0 \frac{t^2}{(t^2+1)(t^2+2at+1)}\,dt
\end{align}
This integral is not very hard to compute, for instance one can do it by partial fraction decomposition or contour integration to get:
\begin{align}
F'(a) =\frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{2\ \sqrt[]{1-a^2}}
\end{align}
We know that:
\begin{align}
I = F(1) = \int^1_0 F'(a)\,da = \frac{1}{2}\int^1_0 \frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{\sqrt[]{1-a^2}}\,da
\end{align}
This looks a bit scary, but hey it is very innocent after setting $a=\cos(x)$, because then one gets:
\begin{align}
I = \frac{1}{2}\int^0_{\pi/2} \frac{\arctan\left(\tan(x)\right)}{\sin(x)}(-\sin(x))\,dx = \frac{1}{2}\int^{\pi/2}_0 x\,dx  = \frac{\pi^2}{16}
\end{align}
