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?

Split the integral at $\frac{1}{\sqrt{2}}$ and use the substitution $x = \sqrt{1-y^2}$ in the second part to obtain
\begin{align}
I &\equiv \int \limits_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_{\frac{1}{\sqrt{2}}}^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x \\
&= \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{y \ln(y+\sqrt{1-y^2})}{1-y^2} \, \mathrm{d} y = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{ \ln(x+\sqrt{1-x^2})}{x(1-x^2)} \, \mathrm{d} x \, .
\end{align}
Now let $x = \sin (t/2)$ to find
\begin{align}
I &= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)}{\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left[\left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)^2\right]}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t \\ 
&= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)\right)}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\sin(t)\right)}{\sin(t)} \, \mathrm{d} t \\
&=\frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(t)\right)}{\cos(t)} \, \mathrm{d} t \, .
\end{align}
Define (idea from this question)
$$ f(a) \equiv \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(a)\cos(t)\right)}{\cos(t)} \, \mathrm{d} t $$
for $ a \in [0,\frac{\pi}{2}]$ and observe that $f(0)=I$ and $f(\frac{\pi}{2}) = 0$. Compute (using $\tan(\frac{t}{2}) = s$)
\begin{align}
f'(a) &= - \frac{\sin(a)}{2} \int \limits_0^{\frac{\pi}{2}} \frac{1}{1+\cos(a)\cos(t)} \, \mathrm{d} t = - \sin(a) \int \limits_0^1 \frac{\mathrm{d} s}{1+\cos(a) + (1-\cos(a))s^2} \\
&= - \frac{\sin(a)}{1+\cos(a)} \sqrt{\frac{1+\cos(a)}{1-\cos(a)}} \arctan \left(\sqrt{\frac{1-\cos(a)}{1+\cos(a)}}\right) \\
&= - \frac{\sin(a)}{\sqrt{1-\cos^2 (a)}} \arctan\left(\tan\left(\frac{a}{2}\right)\right) = - \frac{a}{2} \, .
\end{align}
And finally,
$$ I = f(0) = f \left(\frac{\pi}{2}\right) + \int \limits_{\frac{\pi}{2}}^0 f'(a) \, \mathrm{d} a = 0 + \int \limits_0^{\frac{\pi}{2}} \frac{a}{2} \, \mathrm{d} a = \frac{\pi^2}{16} \, .$$
