Easier ways to prove $\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$
Solution 1:
Note that $f(x)=x^x$ is convex in $[0,1]$ with its minimum happening at $1/e$. Now since $2^e>e$ ($e>1$) it holds that $f(1/e)>1/2$. Therefore \begin{align} \int_0^1 \frac{\log^2 x-2}{x^x}dx<2\int_0^1 (\log^2 x-2)dx=0 \end{align}