Chọn A

Đặt $t={{x}^{x}}$. Khi $\left\{ \begin{align}  & x=1\Rightarrow t=1 \\  & x=e\Rightarrow t={{e}^{e}} \\ \end{align} \right.$

Ta có:

$\begin{align}  & \ln t=x\ln x,\left( \ln t \right)'=\left( \ln x+1 \right)dx\Leftrightarrow \frac{dt}{t}=\left( 1+\ln x \right)dx \\  & \Rightarrow dt=t\left( 1+\ln x \right)dx={{x}^{x}}\left( 1+\ln x \right)dx \\ \end{align}$

Vậy $\int_{1}^{e}{{{x}^{x}}\left( 1+\ln x \right)dx}=\int_{1}^{{{e}^{e}}}{dt}={{e}^{e}}-1$