$\frac{6}{x+6}+\frac{1}{x-2}=\frac{8}{x^2+4x-12}$

$\int e^{-2t}dt$

$\lim_{x\to-4}\left(\frac{x^4-2x^3-19x^2+32x+48}{x^2-16}\right)$

$\frac{x^3+6x^2+7x+2}{x+2}$

$\left(3n+4\right)^3$

$\lim_{x\to\infty}\left(x^2+x\right)e^{-x}$

$\int\left(x^4e^{\frac{1}{2}x}\right)dx$