$\int_{-\infty}^{-1}\left(\frac{1}{\left(x^2-6x\right)^2}\right)dx$
$\lim_{x\to0}\left(\frac{x^2}{x^3-x^2}\right)$
$\frac{dy}{dx}=\left(y-2\right)\left(x^2+1\right)$
$\frac{-m}{4m+36}=\frac{1}{m+8}$
$x-2\ge-5$
$2x^2+7+6x-1-7x-6x^2+1$
$x^5=\cot\left(y\right)$
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