$x\left(3x-1\right)^3$
$\left(4d+2b\right)\left(-mn\right)$
$\frac{dy}{dx}=\frac{-x}{\sqrt{x^2+9}}$
$-597+-78$
$2p^5+6q^42p^5-6q^4$
$\frac{x+4}{3}+3\left(x+2\right)=\frac{4x}{2}+2$
$\int\left(1+\sqrt[2]{\tan\left(x\right)}\right)^2\sec^2\left(x\right)dx$
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