Relatively recently I came across an integral on some forum in this post
Complex contour solution
The target integral can be convolved into a more compact form:
$$\int_{0}^{\pi}\frac{\arctan\left(\frac{\ln\left(\sin x\right)}{x}\right)}{\ln^{2}\left(x^{2}+\ln^{2}\left(\sin x\right)\right)+4\arctan^{2}\left(\frac{\ln\left(\sin x\right)}{x}\right)}dx = \dfrac{1}{4} \operatorname{Im} \int_{0}^{\pi}\frac{dx}{\ln\left(x+i\ln\left(\sin x\right)\right)} =: \dfrac{1}{4} \operatorname{Im} I$$
Using the substitution $z = x+i\ln \left( \sin x \right)$ we realize that there is no derivative $1+i \cot x$ in the integral $I$
This problem can be circumvented in the following way:$$\begin{align}&1+i\cot x\\=& i\left(-i+\cot x\right)\\=& i\left(i+\cot x-2i\right)\\=& i\left(i+\frac{\cos x}{\sin x}-2i\right)\\=& i\left(\frac{\cos x+i\sin x}{\sin x}-2i\right)\\=&i\left(\frac{e^{ix}}{\sin x}-2i\right)\\=&i\left(e^{-\ln\left(\sin x\right)+ix}-2i\right)\\=&i\left(e^{i\left(x+i\ln\left(\sin x\right)\right)}-2i\right)\end{align}$$
Thus:
$$\frac{1+i\cot x}{i\left(e^{i\left(x+i\ln\left(\sin x\right)\right)}-2i\right)}=1 \quad$$implies$$ i=\frac{1+i\cot x}{e^{i\left(x+i\ln\left(\sin x\right)\right)}-2i}\ \ \forall x \in \mathbb{R}$$
and
$$\begin{align}&\int_{0}^{\pi}\frac{dx}{\ln\left(x+i\ln\left(\sin x\right)\right)}\\=&\frac{1}{i}\int_{0}^{\pi}\frac{1}{\ln\left(x+i\ln\left(\sin x\right)\right)}\cdot i\cdot dx\\=&\frac{1}{i}\int_{0}^{\pi}\frac{1}{\ln\left(x+i\ln\left(\sin x\right)\right)}\frac{1+i\cot x}{e^{i\left(x+i\ln\left(\sin x\right)\right)}-2i}dx\\=&\frac{1}{i}\int_\Gamma \dfrac{1}{\ln z} \dfrac{dz}{e^{iz}-2i}\end{align}$$
Next step is solving contour integral constucted by $\Gamma=x+i\ln (\sin x)$ and $L=x+i\ln (\sin a)$, $x\in [a, \pi-a]$ and $a$ will go to $0$
The contour sketch (red elements are "singular points"):
It's easy to show, what $\oint_C=-\frac{\pi i}{\ln\left(\frac{\pi}{2}-i\ln2\right)}$, $\ \int_L \to 0\ $ and $\ \int_\Gamma=\frac{\pi}{\ln\left(\frac{\pi}{2}-i\ln2\right)}$
After all we have:$$\int_{0}^{\pi}\frac{dx}{\ln\left(x+i\ln\left(\sin x\right)\right)}=\frac{\pi}{\ln\left(\frac{\pi}{2}-i\ln2\right)}$$
Taking imaginary parts of equation and multiply both parts by $-\frac{1}{4}$:$$\int_{0}^{\pi}\frac{\arctan\left(\frac{\ln\left(\sin x\right)}{x}\right)}{\ln^{2}\left(x^{2}+\ln^{2}\left(\sin x\right)\right)+4\arctan^{2}\left(\frac{\ln\left(\sin x\right)}{x}\right)}dx=\frac{-\pi\arctan\left(\frac{2\ln2}{\pi}\right)}{\ln^{2}\left(\frac{\pi^{2}}{4}+\ln^{2}2\right)+4\arctan^{2}\left(\frac{2\ln2}{\pi}\right)}$$
Attempt to the real solution
The structure of integrand in the target integral is $\frac{b}{a^2+b^2}$
I tried use $\int_{0}^{\infty}e^{-at}\sin btdt=\frac{b}{a^{2}+b^{2}}$, but inner integral doesn't bounded in some area:
Is there some other real approach to calculating this integral?