$$\int_{0}^{\pi/2}{1+2\cos{(2x)}\ln{(\tan{x})}\over 1+\tan{x}}\mathrm dx=-{\pi\over 4}\tag1$$
Recall: $cos(2x)=\cos^2{x}-\sin^2{x}$
$$\int_{0}^{\infty}{(1+\ln{\tan^2{x}})\cos^2{x}+(1-\ln{\tan^2{x}})\sin^2{x}\over 1+\tan{x}}\mathrm dx$$
Enforcing $u=\tan^2{x}$ then $du=2\tan{x}\sec^2{x}dx$
Recall: $1+\tan^2{x}=\sec^2{x}$ and $1+\cot^2{x}=\csc^2{x}$
$${1\over2}\int_{0}^{\infty}{1+\ln{u}+u(1-\ln{u})\over 1+u}\cdot{\mathrm du\over u^{1/2}+u}$$
I am stuck! Help to prove $(1)$
If we set $$ l(x) = \frac{1+2\cos(2x)\log\tan(x)}{1+\tan(x)}\tag{1} $$ we have: $$ l(x)+l\left(\frac{\pi}{2}-x\right)=1+2\cos(2x)\log\tan x \tag{2}$$ hence the original integral equals $$ I=\frac{\pi}{4}+\int_{0}^{\pi/2}(2\cos^2(x)-1)\log\tan(x)\,dx = \frac{\pi}{4}+\int_{0}^{+\infty}\frac{(1-t^2)\log t}{(1+t^2)^2}\tag{3} $$ and we may exploit symmetry again. By setting $m(t)=\frac{(1-t^2)\log t}{(1+t^2)^2}$, we have $ m(t) = \frac{1}{t^2}m\left(\frac{1}{t}\right) $, so: $$ I = \frac{\pi}{4}+2\int_{0}^{1}\sum_{n\geq 0}(-1)^n (2n+1) t^{2n}\log(t)\,dt = \frac{\pi}{4}-2\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\color{red}{-\frac{\pi}{4}}.\tag{4} $$