Easier ways to prove $\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$ Prove that
$$\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$$
One way to do this is use the idea in the proof of Sophomore's dream. We have
$$x^{-x}=\exp(-x\log x)=\sum_{n=0}^\infty\frac{(-1)^nx^n\log^n x}{n!}$$
Therefore, using the change of variable $x=\exp(-t/(n+1))$ we have
$$\begin{aligned}\int_0^1 \frac{\log^2 x}{x^x}dx=&\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1x^n\log^{n+2}xdx\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}}{n!}\int_0^\infty t^{n+2}e^{-t}dt\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}(n+2)!}{n!}\\
=&\sum_{n=0}^\infty(n+1)^{-(n+2)}(n+2)\\
=&\sum_{n=1}^\infty n^{-(n+1)}(n+1)\\
=&\sum_{n=1}^\infty n^{-n}+n^{-(n+1)}\\
<&2\sum_{n=1}^\infty n^{-n}\\
=&\int_0^1\frac{2}{x^x}dx
\end{aligned}$$
Hence the result follows.
I am curious if there are any other methods to prove this, especially I am interested in easier approaches.
P.S. This was a bonus problem in an assignment from a multivariable calculus class. 
 A: 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}
A: Using the inequality $\log(x)<x-1$ for all $x>0$ we have
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx} 
\end{align}
Also we use the inequality $e^{x} \ge 1+x$ for all $x>0$ to express the denumeantor 
$$x^x=\exp(x\ln(x))\ge 1+x\ln(x).$$ 
Therefore, 
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx} <\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\ln(x)}}dx}.
\end{align}
Now, using the inequality $\log(x) > \frac{x-1}{x}$ for all $x > 0$, we egt
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx} &<\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\ln(x)}}dx}
\\
&<\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\cdot \frac{x-1}{x}}}dx}
\\
&<\int_0^1 {e\cdot x\cdot \frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\cdot \frac{x-1}{x}}}dx} \,\,\,\,\,\,\,\,\,\,\,\,(\text{since $0<x<1$})
\\
&<\int_0^1 {e\cdot \left({\left( {x - 1} \right)^2  - 2}\right)dx} = -\frac{5}{3} e <0
\end{align}
