Why is $\int_0^1 x^x \ln(x) \space dx = -\int_0^1 x^x \space dx$

Question

Why is $\int_0^1 x^x \cdot \ln(x) \space dx = -\int_0^1 x^x \space dx$

If I enter these two integrals in Wolfram Alpha I numerically get the same result, but I couldn't find out why (I tried out some substitutions and integration by parts).

I've come across this problem in an exercise where I have to prove that $\int_0^1\int_0^1 (xy)^{xy} \space dx \space dy = \int_0^1 x^{x} \space dx$

Answer 1

Note that $$(x^x)' = x^x + x^x\ln x $$ Then, with $\int_0^1 d(x^x)=0$

$$\int_0^1 x^x \ln x \space dx = -\int_0^1 x^x \space dx $$