$\frac{sec^2\left(x\right)-1}{\tan^2\left(x\right)+1}$
$16a^2\:x^4-25b^4\:y^2$
$y''+2y'+2y=0$
$2+\frac{1}{\sqrt{n}}$
$3x-9x+1$
$\frac{5x^2-8x-x^3-50}{x+4}$
$\int\left(5t-2\right)^{14}dt$
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