Proof the following $$\frac{2}{3}\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx$$
I've tried to use x=tan(u) substitution, but don't know how to proceed with this $$\frac{2}{3}\int_0^{π/4}\ln(\tan(u))du=\int_0^{π/12}\ln(\tan(u))du$$
Please help
Utilize the integral $$\int_0^1 \frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dy =\ln(\tan t) $$ to prove
\begin{align} &\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx \\= & \int_0^\frac\pi{12}\ln(\tan t)~dt = \int_0^1 \int_0^{\frac\pi{12} } \frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dt \ dy\\= &\ -\frac12\int_0^1\frac1y \tan^{-1}\frac y{1-y^2} dy = -\frac12\int_0^1\frac{\tan^{-1}y^3}y \overset{y^3\to y}{dy} -\frac12\int_0^1\frac{\tan^{-1}y}y dy\\=&- \frac23\int_0^1\frac{\tan^{-1}y}ydy \overset{ibp} =\frac23\int_0^1\frac{\ln y}{1+y^2}dy \end{align}
