$y = \frac { 6 x + 2 } { 5 x - 1 }$

$\frac{x^3+5x^2+6x+12}{x+4}$

$\int\tan^6\left(x\right)\sec^8\left(x\right)dx$

$f=\:\left(1\:-\:i\right)^6+\:\left(1\:+\:i\right)^2$

$\left(9x^4\:-100y^2\right)\:\left(9x^4\:+100y^2\right)$

$\left(1-sin^2\theta\:\right)\left(1+tan^2\theta\:\right)$

$7x^3+3x+7x-3x^3+3x^3-x^3$