$\int\tan^2\left(6x-3\right)dx$
$4x+7y-3x+5+3y-x-5$
$\int e^{10}dx$
$\lim_{x\to0}\left(\frac{e^{x}-1}{x^{3}}\right)$
$-5\::\:-20$
$\left(4x^4+3x^3-x^2\right)\left(x^2-x\right)$
$\frac{dy}{dx}+y=\frac{1}{3}$
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