Evaluating $\int x^x\ln(xe) \,\mathrm dx $

Question

$$\int x^x\ln(xe) \,\mathrm dx $$ we got this problem so i seperated this as $\int x^x \ln(x) \,\mathrm dx $ + $\int x^x \ln(e) \,\mathrm dx $ it becomes $\int x^x \ln(x) \,\mathrm dx$ + $\int x^x \,\mathrm dx $ using $ \ln(ab) = \ln(a)+\ln(b) $ now what to do ? $x^x$ integration seems very difficult and so byparts integral can't be done in here , so so what to do ?

Answer 1

$Hint: \frac{ \mathrm d}{\mathrm dx}(x^{x})=x^{x}(\ln(x)+1)$

Answer 2

HINT:

$$\int x^x\ln(xe) \,\mathrm dx=\int (e^{\ln x})^x (\ln x+\ln e ) \,\mathrm dx =\int e^{x\cdot\ln x}(\ln x+1) \,\mathrm dx$$

Put $x\cdot\ln x=u$