$I=\int_0^{\frac{\pi}{4}} \dfrac{1}{\cos^2 x ( 2\tan x + 1)^2 }dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{1}{\sin^2 x (2+\cot^2 x)^2}dx$
$=\int \dfrac{1}{(2\tan x +1)^2} d(\tan x) -\int \dfrac{1}{(\cot x +2)2} d(\cot x)$
$=\int_0^1 \dfrac{1}{(2t+1)^2}dt - \int_1^0 \dfrac{1}{(t+2)^2}dt $ Dễ rồi tự làm nốt