Ta có \(I=\int_{0}^{3}3x^2dx+\int_{0}^{3}3x\sqrt{x^2+16}dx\) \(I_1=\int_{0}^{3}3x^2dx=x^3\bigg|^3_0=27\) \(I_2=\int_{0}^{3}3x\sqrt{x^2+16}dx\) Đặt \(t=x^2+16\), ta có \(t'=2x;t(0)=16,t(3)=25\) Do đó \(I_2=\int_{16}^{25}\frac{3}{2}\sqrt{t}dt\) \(=t\sqrt{t}\bigg|^{25}_{16}=61\) Vậy \(I=I_1+I_2=88\)