\(=\frac{1}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\ln\left(\tan x\right)d\left[\ln\left(\tan x\right)\right]=\frac{1}{4}\left[\ln^2\left(\tan x\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{4}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{16}\ln^23\)
Đặt \(t=\tan x\Rightarrow\begin{cases}dt=\frac{dt}{\cos^2}=\left(1+t^2\right)dx\rightarrow dx=\frac{dt}{1+t^2}\\x=\frac{\pi}{4}\rightarrow t=1;x=\frac{\pi}{3}\rightarrow t=\sqrt{3}\end{cases}\)
Khi đó : \(I=\int\limits^{\sqrt{3}}_1\frac{\ln t}{\frac{2t}{1+t^2}}.\frac{dt}{1+t^2}=\frac{1}{2}\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\frac{1}{2}J\left(1\right)\)
\(J=\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\int\limits^{\sqrt{3}}_1\ln.d\left(\ln t\right)=\frac{1}{2}\ln^2t|^{\sqrt{3}}_1=\frac{1}{2}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{8}\ln^23\)
Thay vào (1) ta có : \(I=\frac{1}{16}\ln^23\)
