Start from the opposite task.
If $\displaystyle \int x^x \, dx=F(x)$ then $\displaystyle F'(x)=x^x$
First we need to find asymptotic evaluation of the integral. Let us take it in the form
$F(x)=x^xg(x)$
So it has to be:
$F'(x)=x^x((1+\ln(x))g(x)+g'(x))=x^x$
From there it is sufficient to take $g(x) \sim \frac{1}{1+\ln(x)}$
So we can start our journey:
$F(x)=x^x(\frac{1}{1+\ln(x)}+f(g(x)))$
If you calculate the derivative of this you have
$g'(x)f'(g(x))-\frac{1}{x(\ln(x)+1)^2}+(\ln(x)+1)f(g(x))=0$
For the purpose of cancellation if it best to take
$g'(x)=\ln(x)+1$
meaning
$g(x)=x\ln(x)$
Now we continue using the steps that are revealing the integral structure.
$F(x)=x^x(\frac{1}{\ln(x)+1}+f(x\ln(x)))$
Take derivative once more and you have got $f(x\ln(x))=\frac{1}{x(1+\ln(x))^3}-f'(x\ln(x))$ or $F(x)=x^x(\frac{1}{\ln(x)+1}+\frac{1}{x(\ln(x)+1)^3}-f'(x\ln(x)))$
We can then write:
$F(x)=x^x(\frac{1}{\ln(x)+1}+\sum_{n=1}^{\infty}f_n(x))$
where
$\displaystyle f_{n}=-\frac{f_{n-1}'}{1+\ln(x)},\,f_0=\frac{1}{1+\ln(x)}$
From $x=0$ to $1$ you can use probably more suitably $F(x)=xg(\ln(x))$
The derivation is similar to the one given above.