$\int\frac{\left(4x-1\right)}{\left(x+1\right)\left(x-4\right)\left(x+4\right)}dx$
$\left(x+3\right)^2<2$
$\log\left(10x\sqrt[3]{10}\right)$
$-2x\:-\:8\:-\:7x\:+\:2$
$x^3\left(7.5\right)x$
$\int te^{9t}dt$
$\frac{dx}{dy\:}=\frac{y}{2x}$
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