$sec\:u\:+tan\:u\:=\frac{cos\:u}{1-sin\:u}$

$\int\left(1+2x\right)^4\left(2x\right)dx$

$\lim\:_{x\to\:\infty\:}\left(cos\left(\frac{6}{x}\right)\right)^2$

$-x^2-2x+2x^2$

$\frac{x}{3}-\frac{\left(x+2\right)}{5}=2$

$4+13+1+-8$

$z^2-14z$