$3x^2-1-x^2+3x$
$\int\frac{x^2+3}{\left(x^3-1\right)\left(x+2\right)}dx$
$\frac{d^2}{dx^2}\left(3x^3-x^2+6x-4\right)$
$\frac{dy}{dx}=\frac{\tan\left(5x\right)sec^3\left(5x\right)}{sen^2\left(9y\right)cos\left(9y\right)}$
$5x^2-6x+9$
$2x^3-7x$
$\frac{1}{7}\cdot7$
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