If ∫ _ 0 ^ 3 15 x ^ 3 1 + x ^ 2 + (1 + x ^ 2 ) ^ 3 dx = α 2 + β 3 ,...

Given,

${∫}_{0}^{\sqrt{3}}\dfrac{15{x}^{3}}{\sqrt{1 + {x}^{2} + \sqrt{{\left(1 + {x}^{2}\right)}^{3}}}}dx = \alpha\sqrt{2} + \beta\sqrt{3}$,

Taking L.H.S we get ${∫}_{0}^{\sqrt{3}}\dfrac{15{x}^{3}}{\sqrt{1 + {x}^{2} + \sqrt{{\left(1 + {x}^{2}\right)}^{3}}}}dx$

Now put $1 + {x}^{2} = {t}^{2}$

$⇒ 2xdx = 2tdt$

$⇒ xdt = tdt$

$∴ {∫}_{1}^{2}\dfrac{15\left({t}^{2} - 1\right)tdt}{\sqrt{{t}^{2} + {t}^{3}}}$

$= 15{∫}_{1}^{2}\dfrac{t\left({t}^{2} - 1\right)}{t\sqrt{1 + t}}dt$

Now put $1 + t = {u}^{2}$$⇒ dt = 2udu$

So, the integral will be $15{∫}_{\sqrt{2}}^{\sqrt{3}}\dfrac{{\left({u}^{2} - 1\right)}^{2} - 1}{u} \times 2udu$

$= 30{∫}_{\sqrt{2}}^{\sqrt{3}}\left({u}^{4} - 2{u}^{2}\right)du$

$= 30{\left(\dfrac{{u}^{5}}{5} - \dfrac{2{u}^{3}}{3}\right)}_{\sqrt{2}}^{\sqrt{3}}$

$= 30\left[\dfrac{1}{5}\left({\sqrt{3}}^{5} - {\sqrt{2}}^{5}\right) - \dfrac{2}{3}\left({\sqrt{3}}^{3} - {\sqrt{2}}^{3}\right)\right]$

$= 30\left[\dfrac{1}{5}\left(9\sqrt{3} - 4\sqrt{2}\right) - \dfrac{2}{3}\left(3\sqrt{3} - 2\sqrt{2}\right)\right]$

$= 30\left[- \dfrac{1}{5} \times \sqrt{3} + \dfrac{8}{15}\sqrt{2}\right]$

$= - 6\sqrt{3} + 16\sqrt{2}$

So, ${∫}_{0}^{\sqrt{4}}\dfrac{15{x}^{3}}{\sqrt{1 + {x}^{2} + \sqrt{{\left(1 + {x}^{2}\right)}^{3}}}}dx = \alpha\sqrt{2} + \beta\sqrt{3}$

$⇒ - 6\sqrt{3} + 16\sqrt{2} = \alpha\sqrt{2} + \beta\sqrt{3}$

Now on comparing we get, $\alpha = 16, \beta = - 6$

$∴ \alpha + \beta = 10$