Ta tách tích phân I như sau: \(I=\int_{1}^{2}xdx+\int_{1}^{2}\frac{ln^2x}{x}dx\) \(I_1=\int_{1}^{2}xdx=\frac{x^2}{2}\bigg|^2_1=\frac{3}{2}\) \(I_2=\int_{1}^{2}\frac{ln^2x}{x}dx\) Đặt \(t=lnx\Rightarrow dt=\frac{1}{x}dx\) Đổi cận: \(x=2\Rightarrow t=ln2;x=1\Rightarrow t=0\) \(I_2=\int_{0}^{ln2}t^2dt=\frac{t^3}{3}\bigg |^{ln2}_0=\frac{ln^32}{3}\) Vậy \(I=I_1+I_2=\frac{3}{2}+\frac{ln^32}{3}\)