$20x+40=40x+20$
$-15+16+4-19+1$
$4-7-13-16-2$
$\int\left(\frac{1}{x^2+6x+10}\right)dx$
$\int5\left(\arctan\left(\frac{x}{2}\right)\right)dx$
$\left(-4\right)\left(-5\right)\left(-3\right)$
$\int\frac{-x+1}{x^2\left(3x^2+4\right)}dx$
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