$\int\:\frac{3w^5+6w^4-48w^3-380w-112}{w^3-16w}dw$
$\int tan^2\frac{5x}{3}sec^4\frac{5x}{3}dx$
$\frac{dy}{dx}=\frac{x+2}{2y}$
$x\left(5x+x^2\right)$
$\frac{dx}{dx}=ax^2+1$
$\left(a^5\right)^3$
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