$\int\frac{3x+8}{x\left(x+2\right)^2}dx$
$x-14=-21$
$\left(x^{\frac{3}{2}}\right)\left(x^{-\frac{1}{2}}\right)$
$\int costdt$
$\lim_{x\to\infty}\left(8x^2+4x-5\right)$
$\frac{x^3-6x^2+3x+10}{x+1}$
$\frac{\left(20x^4+4x^3-2x^2+3x-8\right)}{5x^2+1x-3}$
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