$\int\frac{x+12}{\left(x-6\right)^2\left(x+3\right)}dx$
$y'=xy-2y+x-2$
$\int x^3\:\sqrt{81-x^2}dx$
$\frac{12+r-r^2}{r^3+3r^3}$
$\frac{1}{3}\left(2\cdot6\right)-\frac{3}{4}\left(6^2\right)$
$\left(-5\right)^2-\left(-12\right)$
$1-2\sin^2\infty=2\cos^2\infty-1$
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