I've seen many many questions on the internet with answer that it cannot be done with elementary functions. Now I did this integration myself and got a pretty nice result. Since I've seen so many answers telling it can't be done I have no idea where I might have gone wrong.
$$K=\int x^x \,dx$$ so $$\ln(K)=\ln\int x^x \,dx =\int \ln x^x \,dx = \int x\ln x \,dx\,.$$
Integration of $x\ln x$ can be done relatively simply.
$$\begin{align*} \int x\ln x \,dx &= (\ln x)\frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x} \,dx\\ &= (\ln x)\frac{x^2}{2}-\frac{1}{2}\int x \,dx \\ &= \ln(x)\frac{x^2}{2}-\frac{1}{2}\frac{x^2}{2} \\ &= \ln(x)\frac{x^2}{2}-\frac{x^2}{4} \\ &= \frac{(\ln x^2)x^2-x^2}{4} \\ &= \ln K\end{align*}$$
Thus making $K$ to be equal to $K = e^{\frac{ln(x^2)*x^2-x^2}{4}+c}$
This seems rather elementary function to me, but I may have done some nasty mistake. Where did I go wrong?