$\left(\frac{1}{3}t+3\right)^3$
$\int\sin^2\left(6x\right)\cos^5\left(6x\right)dx$
$-5x-2x-3x$
$-1\:-4\:3\:3$
$-7x^{-5}$
$x\left[\ln\left(\frac{x}{y}\right)+1\right]dy-ydx=0$
$9y^2-16$
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