$\int\frac{1}{\left(x-3\right)\left(x+3\right)}dx$
$x+5<\:4$
$\left(4u+1\right)^5$
$\left(-5+3\right)-\left(-44\right)$
$\left|\left(-28\right)-\left(42\right)\right|-13-\left(-10\right)$
$\int\:x\left(x^2+1\right)^2\:dx$
$\frac{64x^3-8}{4x-2}$
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