Biết $I=\int\limits_{1}^{e}{{{x}^{2}}\ln xdx}=a{{e}^{3}}+b$ với a...

Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dx={{x}^{2}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right.$.
$\begin{aligned}
& \Rightarrow I=\left( \dfrac{{{x}^{3}}}{3}\ln x \right)\left| \begin{aligned}
& ^{e} \\
& _{1} \\
\end{aligned} \right.-\int\limits_{1}^{e}{\left( \dfrac{{{x}^{3}}}{3}.\dfrac{1}{x} \right)dx}=\dfrac{{{e}^{3}}}{3}-\dfrac{1}{3}\int\limits_{1}^{e}{{{x}^{2}}dx}=\dfrac{{{e}^{3}}}{3}-\dfrac{1}{3}.\dfrac{{{x}^{3}}}{3}\left| \begin{aligned}
& ^{e} \\
& _{1} \\
\end{aligned} \right.=\dfrac{{{e}^{3}}}{3}-\dfrac{{{e}^{3}}}{9}+\dfrac{1}{9}=\dfrac{2{{e}^{3}}}{3}+\dfrac{1}{9} \\
& \Rightarrow a=\dfrac{2}{9},b=\dfrac{1}{9}\Rightarrow 9\left( a+b \right)=3. \\
\end{aligned}$