$\frac{1}{-8+x^3}$
$4x^2-8x=0\:$
$-\int_{-1}^3\left(3x^2+4x-5\right)dx$
$x^2+6x+8\le\:\:0$
$y'=\:x^2\:+5x^2y$
$\frac{-12y}{3}-\frac{18}{3}$
$15-\left(-61\right)$
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