Let f(x) = ∫ dx (3 + 4 x ^ 2 ) 4 - 3 x ^ 2 , |x| 0 , then α ^ 2 + β...

Given,

$f\left(x\right) = ∫ \dfrac{dx}{\left(3 + 4{x}^{2}\right)\sqrt{4 - 3{x}^{2}}}$

Put $x = \dfrac{1}{t}, dx = - \dfrac{1}{{t}^{2}}dt$

So, $f\left(x\right) = ∫ \dfrac{- dt}{{t}^{2}\left(3 + \dfrac{4}{{t}^{2}}\right)\sqrt{4 - \dfrac{3}{{t}^{2}}}}$

$⇒ f\left(x\right) = ∫ \dfrac{- tdt}{\left(3{t}^{2} + 4\right)\sqrt{4{t}^{2} - 3}}$

Now let, $4{t}^{2} - 3 = {\lambda}^{2} ⇒ 8tdt = 2\lambdad\lambda$

$⇒ f\left(x\right) = - ∫ \dfrac{\lambdad\lambda}{4 \cdot \left(3\left(\dfrac{{\lambda}^{2} + 3}{4}\right) + 4\right) \cdot \lambda}$

$⇒ f\left(x\right) = - ∫ \dfrac{d\lambda}{3{\lambda}^{2} + 9 + 16}$

$⇒ f\left(x\right) = - ∫ \dfrac{d\lambda}{3{\lambda}^{2} + 25}$

$⇒ f\left(x\right) = - \dfrac{1}{3} ∫ \dfrac{d\lambda}{{\lambda}^{2} + \dfrac{25}{3}}$

$⇒ f\left(x\right) = - \dfrac{1}{3} \times \dfrac{\sqrt{3}}{5}\tan^{- 1}\left(\dfrac{\sqrt{3}\lambda}{5}\right) + c$

$⇒ f\left(x\right) = - \dfrac{\sqrt{3}}{15}\tan^{- 1}\left(\dfrac{\sqrt{3}\sqrt{\left(4 - 3{x}^{2}\right)}}{5x}\right) + c$

Now using, $f\left(0\right) = 0 ⇒ c = + \dfrac{\sqrt{3}\pi}{30}$

Hence, $f\left(1\right) = \dfrac{- \sqrt{3}}{15}\tan^{- 1}\left(\dfrac{\sqrt{3}}{5}\right) + \dfrac{\sqrt{3}}{15} \times \dfrac{\pi}{2}$

$⇒ f\left(1\right) = \dfrac{- \sqrt{3}}{15}\left(\tan^{- 1}\left(\dfrac{\sqrt{3}}{5}\right) - \dfrac{\pi}{2}\right)$

$⇒ f\left(1\right) = \dfrac{\sqrt{3}}{15}\left(\dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{\sqrt{3}}{5}\right)\right)$

$⇒ f\left(1\right) = \dfrac{\sqrt{3}}{15}\cot^{- 1}\left(\dfrac{\sqrt{3}}{5}\right)$

$⇒ f\left(1\right) = \dfrac{\sqrt{3}}{15}\tan^{- 1}\left(\dfrac{5}{\sqrt{3}}\right)$

$⇒ f\left(1\right) = \dfrac{1}{5\sqrt{3}}\tan^{- 1}\left(\dfrac{5}{\sqrt{3}}\right)$

Now comparing with given value of $f\left(1\right)$ we get, $\alpha = 5 \text{ and } \beta = \sqrt{3}$

$⇒ {\alpha}^{2} + {\beta}^{2} = 28$