The integral (24)/(π) ∫ _ 0 ^ 2 (2 - x ^ 2 )dx (2 + x ^ 2 ) 4 + x ^...

Let, $I = \dfrac{24}{\pi}{∫}_{0}^{\sqrt{2}}\dfrac{2 - {x}^{2}}{\left(2 + {x}^{2}\right)\left(\sqrt{4 + {x}^{4}}\right)}dx$

$= \dfrac{24}{\pi}{∫}_{0}^{\sqrt{2}}\dfrac{2 - {x}^{2}}{{x}^{2}\left(\dfrac{2}{x} + x\right)\sqrt{\left(\dfrac{4}{{x}^{2}} + {x}^{2}\right)}}dx$

$= \dfrac{24}{\pi}{∫}_{0}^{\sqrt{2}}\dfrac{\dfrac{2}{{x}^{2}} - 1}{\left(\dfrac{2}{x} + x\right)\sqrt{{\left(\dfrac{2}{x} + x\right)}^{2} - 4}}dx$

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

$I = - \dfrac{24}{\pi}{∫}_{\infty}^{2\sqrt{2}}\dfrac{dt}{t\sqrt{{t}^{2} - 4}} = - \dfrac{12}{\pi}{∫}_{\infty}^{2\sqrt{2}}\dfrac{2tdt}{{t}^{2}\sqrt{{t}^{2} - 4}}$

Let ${t}^{2} - 4 = {z}^{2}, 2tdt = 2zdz$

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

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