$\int\:\left(4-3x^2\right)^3\:6xdx$

$4y^2\cdot y^2$

$4^8.16^2.64^3$

$6-y^2=0$

$\left(xz^3\right)\left(\frac{x^2y^{-1}}{3z^2}\right)^{-2}$

$\int\frac{1}{\left(9+y^2\right)^{\frac{3}{2}}}dy$

$-\:\left(-8\:+\:2\right)\:\cdot\left(-2\:-1\right)$