$\frac{6x^3+5x^2-19x-10}{1}$
$16x\:+\:11=18x-9$
$x+y-z=896$
$\int_1^{\infty}\left(\frac{x}{3x+1}\right)dx$
$\:3\cdot3\:+\:\left(-\:9\right)\:-\:\left(-2\cdot2\right)\:$
$\int\left(\log\left(\frac{1}{x}\right)\right)x^4dx$
$\left(3x^2+25\right)\left(3x^2+5\right)$
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