The answer can be in the form of two defined constants: $$A = \frac{1}{1^1}+\frac{1}{2^2} + \frac{1}{3^3} + \cdots$$ $$B = \frac{1}{2^2} + \frac{1}{4^4} + \frac{1}{6^6} + \cdots $$
I would highly appreciate it if you included your thought process in the answer itself.
As pointed out by zz20s , the answer to this integral can be seen on the wikipedia page: Sophomore's dream and the page accompanies a proof too
Note: there is no known closed form for $\int x^x \, \mathrm dx$.
But since we have bounds, we can use froggie's approach in this answer.
Recall that: $$\int_0^1 x^x\, \mathrm dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\, \mathrm dx$$
Make the substitution $u = -\log x$
$$\int_0^1 x^x\, \mathrm dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\, \mathrm dx = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_0^\infty e^{u(k+1)}u^k\, \mathrm du$$$$ = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^{u(k+1)}((k+1)u)^k\, \mathrm du$$
Make the substitution $t = (k+1)u$
$$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^tt^k\, \mathrm dt = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1),$$ where $\Gamma$ is the usual Gamma function. Since $\Gamma(k+1) = k!$, the final expression is $$ \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$
I have written a detailed explanation to find the integral of a general expression of the form $f(x) =x^{cx^a} $ from $0$ to $1$ here.
$$\int_{0}^{1} x^{cx^a}\, dx = 1- \frac{c}{(a+1)^2} + \frac{c^2}{(2a+1)^3} - \frac{c}{(3a+1)^4}+\ldots$$
On substituting $c =1$ and $a=1$, we get the desired result.