If ∫ 0 3 15 x 3 1 + x 2 + 1 + x 2 3 d x = α 2 + β 3 , where α , β are integers,…

Given,

${\int }_0^{\sqrt{3}}\frac{15x^3}{\sqrt{1+x^2+\sqrt{{\left(1+x^2\right)}^3}}}dx=\alpha \sqrt{2}+\beta \sqrt{3}$, 

Taking L.H.S we get ${\int }_0^{\sqrt{3}}\frac{15x^3}{\sqrt{1+x^2+\sqrt{{\left(1+x^2\right)}^3}}}dx$

Now put $1+x^2=t^2$

$\Rightarrow 2xdx=2tdt$

$\Rightarrow xdt=td\mathrm{t}$

$\therefore {\int }_1^2\frac{15\left(t^2-1\right)tdt}{\sqrt{t^2+t^3}}$

$=15{\int }_1^2\frac{t\left(t^2-1\right)}{t\sqrt{1+t}}dt$

Now put $1+t=u^2$$\Rightarrow dt=2udu$

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

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

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

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

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

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

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

So, ${\int }_0^{\sqrt{4}}\frac{15x^3}{\sqrt{1+x^2+\sqrt{{\left(1+x^2\right)}^3}}}dx=\alpha \sqrt{2}+\beta \sqrt{3}$

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

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

$\therefore \alpha +\beta =10$