$\left(3x^2+1\right)y'-2xy=6x$
$\int\frac{8x}{x^3+8}dx$
$-2\cdot\infty^2+1$
$z^2-16z$
$\frac{12q^{-6}}{4\left(q\right)^2}$
$\int\:te^{6t}dt$
$a^2-4b\:=\:1$
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