Không khó lắm đâu
$I=-\dfrac{1}{9} \int_1^{\sqrt 3} x^2.\dfrac{\sqrt{9+3x^2}}{x} . \dfrac{-9 dx}{x^3}$
Đặt $\dfrac{\sqrt{9+3x^2}}{x}=t\Rightarrow \dfrac{9+3x^2}{x^2}=t^2 \Rightarrow \dfrac{9}{x^2}+3 =t^2$
$\Rightarrow -\dfrac{9dx}{x^3}=tdt$
Vậy $I=\dfrac{1}{9} \int \limits_{\sqrt 6}^{2\sqrt 3} \dfrac{9}{t^2-3}. t^2 dt = \dfrac{1}{9} \int \limits_{\sqrt 6}^{2\sqrt 3} \bigg (9+\dfrac{27}{t^2-3} \bigg )dt$
Giờ thì ok rồi nhé, cái $\int \dfrac{1}{t^2-3}dt =\dfrac{1}{2\sqrt 3}\int \bigg (\dfrac{1}{t-\sqrt 3}-\dfrac{1}{t+\sqrt 3} \bigg )dt$