How about this?

Let $I(a,b)=\int x^a\ln (x{)}^b\mathrm{dx}$

(By the way all integrals here are with respect to $x$ and between 0 and 1.)

Integration by parts reveals that:

$I(a,b)=(x^{a+1}\ln (x{)}^b)/(a+1)-b/(a+1)I(a,b-1)$

Provided $a+1\ge b$ the first term on the right hand side will be zero after the limits are applied, so we can then say:

$I(a,b)=-b/(a+1)I(a,b-1)$

By successive application of this formula we can evaluate $I(n,n)$ (for +ve integral $n$) as:

$I(n,n)=(-n)(-n+1)(-n+2)...(-2)(-1)/(n+1{)}^n\times I(n,0)$

Clearly $I(n,0)=1/(n+1)$

So $I(n,n)=(-1{)}^{n}n!/(n+1{)}^{n+1}$

Now back to the problem.

$\int x^{-x}\mathrm{dx}$ can be written as $\int e^{-x\ln x}\mathrm{dx}$ which can now be expanded out using the infinite series. We get

$\int [1-x\ln x+(x\ln x{)}^2/2!-(x\ln x{)}^3/3!+...]\mathrm{dx}$

$=I(0,0)-I(1,1)+I(2,2)/2!-I(3,3)/3!+...$

Now the factorials will cancel with the factorials contained within the $I(n,n)$s. The $(-1{)}^n$ will disappear and we are left with: