$\left(4xy-3x^2\right)^2$
$2x+3\left(x+1\right)$
$\int y\left(3-y^2\right)dy$
$4\left(x+1\right)+2\left(x-4\right)$
$\int\frac{\left(z^4+3\right)}{\left(z+1\right)\left(z^2+1\right)}dx$
$\lim_{x\to0}\left(\frac{4}{x^2}\right)$
$\frac{5}{x-5}=\frac{x}{x-5}$
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