Tính tích phân từ 1 đến e x^2ln(x) d(x)

Đáp án đúng là D

Đặt u = lnx => du = $\frac{1}{\mathrm{x}}$dx

dv = x²dx => v = $\frac{{\mathrm{x}}^3}{3}$+ C

Chọn C = 0 => v = $\frac{{\mathrm{x}}^3}{3}$

$$\begin{gathered} \Rightarrow \underset{1}{\overset{\mathrm{e}}{\int }}{\mathrm{x}}^2\ln \left(\mathrm{x}\right)\mathrm{d}\left(\mathrm{x}\right)={\left.\ln \left(\mathrm{x}\right).\frac{{\mathrm{x}}^3}{3}\right|}_1^{\mathrm{e}}-\underset{1}{\overset{\mathrm{e}}{\int }}\frac{{\mathrm{x}}^3}{3}.\frac{1}{\mathrm{x}}\mathrm{d}\text{x} \\ =\frac{{\mathrm{e}}^3}{3}-{\left.\frac{{\mathrm{x}}^3}{9}\right|}_1^{\mathrm{e}} \\ =\frac{{\mathrm{e}}^3}{3}-\frac{{\mathrm{e}}^3}{9}+\frac{1}{9} \\ =\frac{2{\mathrm{e}}^3+1}{9} \end{gathered}$$