$I=-\int_1^2 \dfrac{-x\sqrt{4-x^2}}{x^2}dx$ đặt $\sqrt{4-x^2}=t \Rightarrow -xdx=tdt$
Vậy $I=-\int \limits_{\sqrt 3}^{0} \dfrac{t^2}{4-t^2}dt=\int_0^{\sqrt 3} \dfrac{t^2 -4 +4}{4-t^2}dt=-\int_0^{\sqrt 3} dt +\int_0^{\sqrt 3} \dfrac{4}{4-t^2}dt$
Cái $I_1 =\int \dfrac{4}{4-t^2}dt =\int \bigg (\dfrac{1}{2+t}+\dfrac{1}{2-t} \bigg )dt =\ln \bigg | \dfrac{t+2}{2-t} \bigg |+C$
Tự lắp lại nhé