$\int x^x\ln \left(ex\right)dx$ is equal to 

$I=\int x^x\ln \left(ex\right)dx=\int x^x(1+\ln x)dx$

Let $t=x^x=e^{x\ln x}\Rightarrow \frac{dt}{dx}=x^x(x.\frac{1}{x}+\ln x)$

$\Rightarrow dt=x^x(1+\ln x)dx\Rightarrow I=\int dt=t+C=x^x+C$