$\lim_{x\to\infty}\left(\frac{\arctan\left(x\right)}{\frac{1}{x}}\right)$
$\int\frac{8}{\sqrt{64-\left(x-9\right)^2}}dx$
$256x^2-196y^2$
$2x^2+17x+35$
$\sqrt{2a}+\sqrt{3b}$
$2^{2x}\:-4\:.\:2^x\:-32=0$
$\left(6a^2-7b\right)\left(6a^2+2b\right)$
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