$\frac{du}{dv}=\frac{\left(3u-2v\right)^2}{\left(2u-3v\right)^2}$
$\int\left(8x\sin\left(2x^2+3\right)\right)dx$
$-459:17$
$-16-\left(-1\right)$
$2\left(3\right)^2\left(8\right)^2$
$\sqrt[3]{4^9}$
$2x+10>5x-4$
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