$\lim_{x\to\infty}\frac{\left(3x-2\right)^2\left(2x+5\right)^4}{\left(6x^6-7x^3+8\right)}$

$x^2-3x-1$

$\frac{x+6}{x-4}=\frac{1}{x+1}$

$\int_0^4\left(x^3\sqrt{x^2+16}\right)dx$

$\lim\:_{x\to\:\infty\:\:}\left(e^{\frac{\sqrt{x^2+1}}{x-1}}\right)$

$\frac{d}{dx}\left(y=x\left(ln8x\right)^2\right)$

$\int_0^2\left(32te^{-4.2t}\right)dt$