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.    

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
$$\int x^x \ln x\,dx=x^x-\int x^x\,dx. $$
