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?

Here is a way to calculate the integral without Feynman's trick or trigonometric substitutions.
By using the substitution previously used by Shashi we have:
$$\underbrace{\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx}_{x=\frac{1}{\sqrt{1+t^2}}}=\underbrace{\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt}_{t=\frac{1}{t}}$$
$$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt-\int _0^{\infty \:}\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt$$
$$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\frac{1}{2}\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$
$$=\frac{1}{2}\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt+\frac{1}{2}\underbrace{\int _1^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt}_{t=\frac{1}{t}}=\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$
$$=\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(1+t^2\right)}{t}\:dt}_{t=t^2}$$
$$=\frac{3}{4}\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt=\frac{3}{4}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}=\frac{3}{8}\zeta \left(2\right)$$
Thus:
$$\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx=\frac{3}{8}\zeta \left(2\right)$$
