$\int\frac{7x-36}{x^2-10x+24}dx$
$\int\frac{\tan^{-1}x}{1+x^2}dx$
$\frac{\left(6x^2+47x+32\right)}{\left(x+7\right)}$
$\left(4x^2+19x-30\right)+\left(9x+36\right)+\left(2x-6\right)$
$\left(-7s^4\right)^{-4}$
$2x-3\le2-3x$
$4\infty$
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