Question

Solve ${\int }_{\sqrt{3}}^{2\sqrt{3}}{x}^3{(x^2-3)}^{3/2}dx=\frac{16038}{35}$.

Answer 1

Let $I={\int }_{\sqrt{3}}^{2\sqrt{3}}{x}^3{(x^2-3)}^{3/2}dx$

Put $x^2-3=t^2$

$\Rightarrow xdx=tdt$
When $x=\sqrt{3}\Rightarrow t=0$

When $x=2\sqrt{3}\Rightarrow t=3$
$I={\int }_0^3(3+t^2)t^{3}tdt$

$={\int }_0^3(3t^4+t^6)dt$
$={[\frac{3t^5}{5}+\frac{t^7}{7}]}_0^3$

$\Rightarrow I=\frac{16038}{35}$