$\int3x^24^{-3x}dx$
$\frac{10k^4}{6k}$
$\frac{d}{dx}\:x^2+\frac{1}{x^2}+3x$
$\left(xy+y^2\right)\cdot y'=y^2$
$3x^2-10x-3\le\:0$
$3x-\frac{4a}{2b^3}$
$3a^2+5ab$
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