$\int\frac{\left(x^3\:-64-12x^2+48x\right)}{x-4}dx$
$\left(-6\right)\cdot\left(+2\right)+\left(+3\right)-\left(6+3-2\right)$
$\int_{-7}^{\infty}\left(e^{-x}\right)dx$
$\sqrt[10]{\frac{\left(3x+4\right)}{2x-4}}$
$\frac{x^2+12x+12}{x\left(x^2-4\right)}$
$5=v+15$
$4g^2+4g-3$
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