$\frac{d}{dx}\left(\frac{1}{\left(t+6\right)\left(t+8\right)\left(t+4\right)}\right)$

$20x+8$

$\lim_{x\to\infty}\left(\frac{1}{x-1}\right)$

$\frac{d^3}{dx^3}\left(x^3+y^3=9\right)$

$\int8.2sen^2\left(1.13+17t\right)dt$

$-6\left(1-4\right)-10$

$\int x^2\left(1+2x^{\frac{3}{2}}\right)^{\frac{2}{3}}dx$