$4+\left(-3+4-3\right)+4-\left(-8+3+2\right)$
$-8\left(2\right)+\left(-5\right)\left(-4\right)$
$\frac{12-3x^2}{4}\ge0$
$\frac{1.1}{2}$
$\int tan^4\left(t\right)dt$
$\int\frac{x+3}{x^2-x+4}dx$
$x^4+6x^2-x^2-30x$
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