$y=arc\:cos\left(\frac{1}{2}x\right)$
$\frac{x-3}{6x-18}+\frac{3-x}{x^2+x-12}$
$\int^6\left(x\cdot\cos\left(2x^2+3\right)\right)dx$
$1\:\frac{-6}{-2}$
$\frac{6}{x+4}=\frac{x}{2}$
$\left|1-x^2\right|\le\left(1+x\right)^2$
$\left(3.4\right)^2$
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