Integral

We substitute $x=\cos y$ , then we have $\int_0^{\frac{\pi}{2}}\frac{\ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy$ make the change of variable $y\mapsto \frac{\pi}{2} -y$ and hence of adding the obtained integral with latter integral, we get $\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy\overset{\text{see Roy's solution}}{=}\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\ln\left(1+\sin y\right)}{\sin y}dy$ Now substitute $y=2\arctan(u)$ and few simplification gives us $\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\ln(1+\sin y)}{\sin y}dy=\int_0^1\left[\frac{2\ln(1+u)-\ln(1+u^2)}{2u}\right]du\\ = \frac{1}{2}\left(-2\operatorname{Li}_2(-1)+\frac{\operatorname{Li}_2(-1)}{2}\right)=-\frac{\pi^2}{12}+\frac{\pi^2}{48}=\frac{\pi^2}{16}$ where $\operatorname{Li}_2(x)$ is dilogarithm function and resultant primitives of last two integrals are due to definition of it.

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