Question

If $\underset{0}{\overset{\sqrt{3}}{\int }}\frac{15x^3}{\sqrt{1+x^2+\sqrt{{\left(1+x^2\right)}^3}}}dx=\alpha \sqrt{2}+\beta \sqrt{3}$, where $\alpha ,\beta$ are integers, then $\alpha +\beta$ is equal to

Answer 1

Answer: 10

Put $1+x^2=t^2$
$2dx=2tdt$
$Xdx=tdt$
$\therefore \underset{1}{\overset{2}{\int }}\frac{15\left(t^2-1\right)tdt}{\sqrt{t^2+t^3}}$
$15\underset{1}{\overset{2}{\int }}\frac{t\left(t^2-1\right)}{t\sqrt{1+t}}dt$
Put $1+t=u^2$
$dt=2udu$
$15\underset{\sqrt{2}}{\overset{\sqrt{3}}{\int }}\frac{{\left(u^2-1\right)}^2-1}{u}\times 2udu$
$30\underset{\sqrt{2}}{\overset{\sqrt{3}}{\int }}\left(u^4-2u^2\right)du$
$30{\left(\frac{u^5}{5}-\frac{2u^3}{3}\right)}_{\sqrt{2}}^{\sqrt{3}}$
$30\left[\frac{1}{5}\left({\sqrt{3}}^5-{\sqrt{2}}^5\right)-\frac{2}{3}\left({\sqrt{3}}^3-{\sqrt{2}}^3\right)\right]$
$30\left[\frac{1}{5}(9\sqrt{3}-4\sqrt{2})-\frac{2}{3}(3\sqrt{3}-2\sqrt{2})\right]$
$30\left[-\frac{1}{5}\times \sqrt{3}+\frac{8}{15}\sqrt{2}\right]$
$-6\sqrt{3}+16\sqrt{2}=\alpha \sqrt{2}+\beta \sqrt{3}$
$\alpha =16,\beta =-6$
$\therefore \alpha +\beta =10$