Indefinite Integral

Please show your working.

I think the answer is x^x(1+lnx).

y=x^x.
logy=xlogx
differentiate
1ydydx=x1x+logx

You can figure out the rest.

Sorry, I did differentiation.

I Think Something Like in this way..........

\hspace{-16}\mathbf{\int x^xdx=\int e^{x.\ln(x)}dx}$\\\\\\ $\mathbf{\int \left(1+\frac{x.\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}+............\right)dx}$\\\\\\ $\mathbf{\int \sum_{n=1}^{\infty}\frac{x^n.\ln^n(x)}{n!}dx}$\\\\\\ $\mathbf{\sum_{n=1}^{\infty}\frac{1}{n!}.\int x^n.\ln^n(x)dx }$

Well I have an approach;

Integrate x^x.1 by parts taking x^x as Ist part and 1 as 2nd term (Is this correct?)

Then probably problem becomes simple!

Check out 63rd question in Arihant Integral Calculus in Exercise of Indefinite Integral! A similar problem!