+ Thứ nhất: Khi tích phân có cận cố định
Đặt $x =\tan t \Rightarrow dx = \dfrac{dt}{\cos^2 t}$
$I =\int_0^{\frac{\pi}{4}} \dfrac{\tan^2 t}{\cos^2 t (1+\tan^2 t)^3}dt = \int_0^{\frac{\pi}{4}} \dfrac{\tan^2 t}{\cos^2 t (\dfrac{1}{\cos^2 t})^3}dt$
$= \int_0^{\frac{\pi}{4}}\sin^2 t \cos^2 t dt = \dfrac{1}{4} \int_0^{\frac{\pi}{4}} \sin^2 2t dt = \dfrac{1}{8} \int_0^{\frac{\pi}{4}} (1-\cos 4t) dt$
$= (\dfrac{1}{8}t -\dfrac{1}{32}\sin 4t ) \bigg |_0^{\frac{\pi}{4}} =\dfrac{\pi}{32}$