$12x^3+10x^2-8x$
$\frac{dy}{dx}=\frac{\left(x+3\right)}{\left(y-3\right)}$
$5\left(u+7\right)$
$\left(\left(m-n\right)-1\right)^2$
$\tan^2x+5=\sec^2x+4$
$\left(2-4x\right)\left(-\frac{7}{6x}\right)$
$\int\left(x^3+3\right)x^3\:dx$
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