Khác quái gì nhau đâu $\tan (x+\dfrac{\pi}{6})=\dfrac{\tan x + \tan \dfrac{\pi}{6}}{1-\tan x \tan \dfrac{\pi}{6}}=\dfrac{\sqrt 3 \tan x+1}{\sqrt 3 -\tan x}$
$I=\int \dfrac{\sqrt 3 \tan^2 x+\tan x}{\sqrt 3 -\tan x}dx$
đặt $\tan x = t \Rightarrow \dfrac{dx}{\cos^2 x}=dt$ hay $(1+\tan^2 x)dx = dt$ $\Rightarrow dx=\dfrac{1}{1+t^2}dt$
Vậy $I=\int \dfrac{\sqrt 3 t^2 +t}{\sqrt 3-t}. \dfrac{1}{t^2+1}dt=\int \bigg (\dfrac{\sqrt 3}{\sqrt 3 -t}-\dfrac{1}{t^2+1} \bigg ) dt$
Dễ rồi nhé