1. We need to evaluate $I=\frac{24}{\pi}\int_0^{\sqrt2}\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,dx.$

Let $J=\int_0^{\sqrt2}\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,dx.$ Then $I=\dfrac{24}{\pi}J$.

2. Use the substitution $x^2=2\tan\theta.$ Then $x=\sqrt{2\tan\theta},$ and differentiating, $2x\,dx=2\sec^2\theta\,d\theta \quad\Rightarrow\quad dx=\frac{\sec^2\theta}{x}\,d\theta.$ Also, $2+x^2=2+2\tan\theta=2(1+\tan\theta),$ $2-x^2=2-2\tan\theta=2(1-\tan\theta),$ and $\sqrt{4+x^4}=\sqrt{4+4\tan^2\theta}=2\sec\theta.$

Now simplify the integrand more systematically.

Since $x=\sqrt{2\tan\theta}$, we have $dx=\frac{\sec^2\theta}{\sqrt{2\tan\theta}}\;d\theta.$ Hence $J=\int \frac{2(1-\tan\theta)}{2(1+\tan\theta)\cdot 2\sec\theta}\cdot \frac{\sec^2\theta}{\sqrt{2\tan\theta}}\,d\theta,$ which looks messy. So instead, use a better substitution.

3. A cleaner substitution is $x=\sqrt2\tan t.$ Then as $x$ goes from $0$ to $\sqrt2$, $t$ goes from $0$ to $\frac\pi4$.

Now compute: $x^2=2\tan^2 t,$ so $2-x^2=2(1-\tan^2 t),$ $2+x^2=2(1+\tan^2 t)=2\sec^2 t,$ $4+x^4=4+4\tan^4 t=4(1+\tan^4 t),$ thus $\sqrt{4+x^4}=2\sqrt{1+\tan^4 t},$ and $dx=\sqrt2\sec^2 t\,dt.$

Therefore, $J=\int_0^{\pi/4}\frac{2(1-\tan^2 t)}{2\sec^2 t\cdot 2\sqrt{1+\tan^4 t}}\cdot \sqrt2\sec^2 t\,dt$ $=\frac{\sqrt2}{2}\int_0^{\pi/4}\frac{1-\tan^2 t}{\sqrt{1+\tan^4 t}}\,dt.$

Now use $1+\tan^4 t=(1+\tan^2 t)^2-2\tan^2 t=\sec^4 t-2\tan^2 t.$ Then $\sqrt{1+\tan^4 t}=\frac{\sqrt{\cos^4 t+\sin^4 t}}{\cos^2 t}.$ So $\frac{1-\tan^2 t}{\sqrt{1+\tan^4 t}}=\frac{\cos2t}{\sqrt{\cos^4 t+\sin^4 t}}.$ Hence $J=\frac{\sqrt2}{2}\int_0^{\pi/4}\frac{\cos2t}{\sqrt{\cos^4 t+\sin^4 t}}\,dt.$

But $\cos^4 t+\sin^4 t=(\sin^2 t+\cos^2 t)^2-2\sin^2 t\cos^2 t=1-\frac12\sin^2 2t.$ So $J=\frac{\sqrt2}{2}\int_0^{\pi/4}\frac{\cos2t}{\sqrt{1-\frac12\sin^2 2t}}\,dt.$

Now let $u=\sin2t \quad\Rightarrow\quad du=2\cos2t\,dt.$ When $t=0$, $u=0$; when $t=\pi/4$, $u=1$. Therefore, $J=\frac{\sqrt2}{4}\int_0^1\frac{du}{\sqrt{1-\frac12u^2}}.$

4. Evaluate the standard integral: $\int\frac{du}{\sqrt{1-a^2u^2}}=\frac{1}{a}\sin^{-1}(au)+C.$ Here, $1-\frac12u^2=1-\left(\frac{u}{\sqrt2}\right)^2.$ So $\int_0^1\frac{du}{\sqrt{1-\frac12u^2}}=\sqrt2\sin^{-1}\left(\frac{u}{\sqrt2}\right)\Bigg|_0^1.$ Thus $=\sqrt2\left[\sin^{-1}\left(\frac1{\sqrt2}\right)-0\right]$ $=\sqrt2\cdot \frac\pi4.$

Therefore, $J=\frac{\sqrt2}{4}\cdot \sqrt2\cdot \frac\pi4=\frac\pi8.$

Finally, $I=\frac{24}{\pi}\cdot\frac\pi8=3.$

5. Final answer: $\boxed{3}$

This matches the stored correct answer.