Từng phần $\left| \begin{align}  & u=\ln x=>du=\frac{1}{x}dx \\  & dv={{x}^{2}}dx=>v=\frac{{{x}^{3}}}{3} \\ \end{align} \right.$

 $\begin{align}  & I=\left( \frac{{{x}^{3}}}{3}\ln x \right)\left| \begin{align}  & e \\ & 1 \\ \end{align} \right.-\int\limits_{1}^{e}{\frac{{{x}^{2}}}{3}dx}=\frac{{{e}^{3}}}{3}-\left( \frac{{{x}^{3}}}{9} \right)\left| \begin{align}  & e \\  & 1 \\ \end{align} \right.=\frac{{{e}^{3}}}{3}-\frac{{{e}^{3}}}{9}+\frac{1}{9} \\  & =\frac{2{{e}^{3}}+1}{9} \\ \end{align}$

Đáp án : A