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

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.

$$\int x^x (\ln x +1-1) dx= \int e^{x\ln x}(\ln x+1)dx -\int x^x dx$$ $$=\int e^{x\ln x} (x\ln x)'dx -\int x^x dx = e^{x\ln x}-\int x^x dx=x^x -\int x^x dx$$ There is now way to solve the last integral.
• Yeah. That's very good approach. Now I can deal with the $\int x^xdx.$ This, I know how to deal with it. Jan 30, 2019 at 21:27
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 $$\int x^x \ln x\,dx=x^x-\int x^x\,dx.$$
• Or you can find $g'(x)$ directly: $(x^x)'=(\exp(x\ln x))'=\exp(x\ln x)\times(x \ln x)'=\exp(x\ln x)(1\times \ln x + \frac1x \times x)=\exp(x\ln x)(\ln x + 1))=x^x(\ln x + 1)$ Jan 31, 2019 at 6:04