Ta có $I=\int\limits_{1}^{e}{x\ln xdx}=\dfrac{a{{e}^{2}}+b}{4}.$...

Đặt$\left\{ \begin{matrix}
u=\ln x\Rightarrow du=\dfrac{1}{x}dx \\
dv=xdx\Rightarrow v=\dfrac{{{x}^{2}}}{2} \\
\end{matrix}. \right.$
$\Rightarrow I=\dfrac{1}{2}{{x}^{2}}\ln x\left| \begin{aligned}
& e \\
& 1 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{1}^{e}{x dx}=\dfrac{1}{2}{{x}^{2}}\ln x\left| \begin{aligned}
& e \\
& 1 \\
\end{aligned} \right.-\dfrac{1}{4}{{x}^{2}}\left| \begin{aligned}
& e \\
& 1 \\
\end{aligned} \right.=\dfrac{1}{2}{{e}^{2}}-\dfrac{1}{4}\left( {{e}^{2}}-1 \right)=\dfrac{1}{4}{{e}^{2}}+\dfrac{1}{4}=\dfrac{{{e}^{2}}+1}{4}$
Vậy $P=a+b+c=1+1=2.$