Evaluate $\int x^x \ln x\, dx$

Question

The integral $$\int x^x \ln x\, dx= ?$$ I know of the integral $\int x^x dx$ can be further simplified as $\int e^{x\ln x} dx$. And this requires identity to simplify. What about the product in the integral $\int x^x\ln x\,dx=\int e^{x\ln x}\ln x\, dx.$ Is there any identity to be used for this one.

Answer 1

$$\int x^x (\ln x +1-1) dx= \int e^{x\ln x}(\ln x+1)dx -\int x^x dx$$ $$=\int e^{x\ln x} (x\ln x)'dx -\int x^x dx = e^{x\ln x}-\int x^x dx=x^x -\int x^x dx$$ There is now way to solve the last integral.

Answer 2

Define $g(x) = x^x$. Then $\ln g(x) = x\ln x$ and differentiating both sides $$\frac{g'(x)}{g(x)}=\ln x+1,$$ which means $g'(x) = x^x(\ln x + 1)$. Now, up to a constant $$x^x = \int g'(x)\,dx = \int x^x \ln x \,dx + \int x^x\,dx$$ thus $$\int x^x \ln x\,dx=x^x-\int x^x\,dx. $$