$\tan\left(x\right)^2=3\sec\left(x\right)-3$
$4x^2-22x-12$
$\frac{18}{\left(\frac{6}{\left(-2\right)}\right)}$
$\frac{d}{dx}\pi\:.\:r^2.h$
$\left(4-x^2\right)\left(x^2+4\right)$
$x-2y+3y+x$
$\int\frac{x^3-3x^2+x-2}{9x^2+18x+8}dx$
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