$\frac{6}{x-2}\ge-4$
$\sqrt{4.18}$
$\int_{-2}^3\left(\frac{48}{x^4}\right)dx$
$\frac{-9+3}{\sqrt{x}-3}$
$\lim_{x\to\infty}\:\left(x+2e^{2x}\right)^{\frac{3}{x}}$
$\lim_{x\to-6}\frac{x+6}{6x^2+14x+48}$
$7\int dx$
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