$\int_0^2\left(6-x\right)dx$
$\frac{d}{dx}\frac{x^2-3x}{3-x}$
$y'=\frac{6x-12}{x^2},y\left(1\right)=20$
$4t^2+8t+4$
$\left(-5\right)\left(-3\right)+\left(-10\right)$
$\ \int\frac{1}{3x\:-1}dx$
$\int130sec^2\left(x\right)dx$
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