$3x-2=1$
$\int_{-\infty}^{-3}\left(\frac{e^{3x}}{1+e^{3x}}\right)dx$
$a^2-49b^{12}$
$\frac{secx\:-\:cscx}{cotx-1}=-\frac{1}{cosx}$
$\int tan^4tsec^4tdt$
$-3x+7>13-5x$
$12.11+18.0+1.013$
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