The integral $$\frac{24}{\pi}\int_0^{\sqrt{2}} \frac{(2-x^2)dx}{(2+x^2)\sqrt{4+x^4}}$$ is equal to ______

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

First, simplify using the substitution $$x^2 = 2\tan\phi$$. Let $$x^2 = 2\tan\phi$$, so $$2x\,dx = 2\sec^2\phi\,d\phi$$, which gives $$x\,dx = \sec^2\phi\,d\phi$$. Since when $$x = 0$$, $$\tan\phi = 0$$ implies $$\phi = 0$$; and when $$x = \sqrt{2}$$, $$\tan\phi = 1$$ implies $$\phi = \frac{\pi}{4}$$.

Next, transform each expression as follows: $$2 - x^2 = 2 - 2\tan\phi = 2(1 - \tan\phi)$$, $$2 + x^2 = 2 + 2\tan\phi = 2(1 + \tan\phi)$$, and $$4 + x^4 = 4 + 4\tan^2\phi = 4(1 + \tan^2\phi) = 4\sec^2\phi$$, so that $$\sqrt{4 + x^4} = 2\sec\phi$$. Also, since $$x = \sqrt{2\tan\phi}$$, it follows that $$dx = \frac{\sec^2\phi}{\sqrt{2\tan\phi}}\,d\phi$$.

Now, substituting into the integral gives $$I = \int_0^{\pi/4} \frac{2(1-\tan\phi)}{2(1+\tan\phi) \cdot 2\sec\phi} \cdot \frac{\sec^2\phi}{\sqrt{2\tan\phi}}\,d\phi = \frac{1}{2\sqrt{2}}\int_0^{\pi/4} \frac{(1-\tan\phi)\sec\phi}{(1+\tan\phi)\sqrt{\tan\phi}}\,d\phi.$$

Since $$\sec\phi = 1/\cos\phi$$ and $$\tan\phi = \sin\phi/\cos\phi$$, this integral becomes $$\frac{1}{2\sqrt{2}}\int_0^{\pi/4} \frac{(\cos\phi - \sin\phi)}{(\cos\phi + \sin\phi) \cdot \sqrt{\sin\phi\cos\phi}}\,d\phi.$$

Now, substituting $$v = \cos\phi + \sin\phi$$ yields $$dv = (\cos\phi - \sin\phi)\,d\phi$$, which matches the numerator times $$d\phi$$. Also, since $$v^2 = 1 + 2\sin\phi\cos\phi$$ we have $$\sin\phi\cos\phi = \frac{v^2 - 1}{2}$$. Moreover, when $$\phi = 0$$, $$v = 1$$, and when $$\phi = \frac{\pi}{4}$$, $$v = \sqrt{2}$$.

Therefore, the integral becomes $$I = \frac{1}{2\sqrt{2}}\int_1^{\sqrt{2}} \frac{dv}{v\sqrt{(v^2-1)/2}} = \frac{1}{2\sqrt{2}} \cdot \sqrt{2}\int_1^{\sqrt{2}} \frac{dv}{v\sqrt{v^2 - 1}} = \frac{1}{2}\int_1^{\sqrt{2}} \frac{dv}{v\sqrt{v^2 - 1}}.$$

Next, using the standard result $$\int \frac{dv}{v\sqrt{v^2 - 1}} = \sec^{-1}|v| + C$$ yields $$I = \frac{1}{2}\Big[\sec^{-1}(v)\Big]_1^{\sqrt{2}} = \frac{1}{2}\left(\sec^{-1}(\sqrt{2}) - \sec^{-1}(1)\right) = \frac{1}{2}\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{8}.$$

Finally, multiplying by the prefactor gives $$\frac{24}{\pi} \cdot I = \frac{24}{\pi} \cdot \frac{\pi}{8} = 3,$$ so that the answer is $$3$$.