$\lim_{x\to1}\left(\frac{\ln\left(x\right)}{x^3-3x\:+2}\right)$
$\int tan\:\left(3x+2\right)\:dx$
$v\left(5\right)=5^2$
$x^2+4x-96=0$
$( - x ^ { 3 } + 125 ) : ( x - 5 )$
$y=x^2+6x+9$
$\left(-4\right)\left(9\right)\left(-2\right)$
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