The integral $\frac{24}{\pi }{\int }_0^{\sqrt{2}}\frac{\left(2-{\mathrm{x}}^2\right)\mathrm{dx}}{\left(2+{\mathrm{x}}^2\right)\sqrt{4+{\mathrm{x}}^4}}$ is equal to ______.

Let, $I=\frac{24}{\pi }{\int }_0^{\sqrt{2}}\frac{2-x^2}{\left(2+x^2\right)\left(\sqrt{4+x^4}\right)}dx$

$=\frac{24}{\pi }{\int }_0^{\sqrt{2}}\frac{2-x^2}{x^2\left(\frac{2}{x}+x\right)\sqrt{\left(\frac{4}{x^2}+x^2\right)}}dx$

$=\frac{24}{\pi }{\int }_0^{\sqrt{2}}\frac{{\displaystyle \frac{2}{x^2}}-1}{\left(\frac{2}{x}+x\right)\sqrt{{\left(\frac{2}{x}+x\right)}^2-4}}dx$

Now let $\frac{2}{x}+x=t, \left(-\frac{2}{x^2}+1\right)dx=dt$

$I=-\frac{24}{\pi }{\int }_{\infty }^{2\sqrt{2}}\frac{{\displaystyle dt}}{t\sqrt{t^2-4}}=-\frac{12}{\pi }{\int }_{\infty }^{2\sqrt{2}}\frac{{\displaystyle 2tdt}}{t^2\sqrt{t^2-4}}$

Let $t^2-4=z^2, 2tdt=2zdz$

$I=-\frac{12}{\pi }{\int }_{\infty }^2\frac{{\displaystyle 2zdz}}{z\left(z^2+4\right)}=-\frac{24}{\pi }{\int }_{\infty }^2\frac{{\displaystyle dz}}{\left(z^2+4\right)}=-\frac{24}{\pi }{\left(\frac{1}{2}{\tan }^{-1}\frac{z}{2}\right)}_{\infty }^2$

$=-\frac{24}{\mathrm{\pi }}\left(\frac{\mathrm{\pi }}{8}-\frac{\mathrm{\pi }}{4}\right)=3$