intregration without substitution of $x^x \ln x$ How do i integrate this without any substitution, purely algebraically :
$$x^x \ln ex$$
I've tried a lot but not have been able to:
$$x^x \ (ln x + 1) = \ln x^{x^x} + x^x$$
or $e^{x \ln x}\ln (x+1)$, i've tried all these methods 
How do proceed after this? 

Let $y=x^x$, then
$$\ln y=x\ln x\implies \frac{y'}{y}=\ln x+1\\ y'=x^x(\ln x+1)=x^x\ln(ex)$$then
$$\int x^x\ln(ex)=y+c=x^x+c$$
