Let $f(x)={\log }_2(x^2+4)$. Then $f$ is increasing on $[0,\infty )$ with $f(0)=2$ and $f(2\sqrt{3})={\log }_2(12+4)=4$. Its inverse function is found by solving $y={\log }_2(x^2+4)$: $x^2=2^y-4$, so $f^{-1}(y)=\sqrt{2^y-4}$ for $y\in [2,4]$.

The two integrals are exactly the area under $f$ on $[0,2\sqrt{3}]$ and the area under $f^{-1}$ on $[2,4]$. By the inverse-function area identity,

${\int }_a^{b}f(x)dx+{\int }_{f(a)}^{f(b)}{f}^{-1}(y)dy=bf(b)-af(a)$

$\alpha =2\sqrt{3}\cdot 4-0\cdot 2=8\sqrt{3}$

$\therefore {\alpha }^2=64\cdot 3=192$.