$\int t\left(arctan\left(t\right)\right)dt$
$\frac{\left(-3\right)\left(+4\right)}{-6}$
$x+1<3$
$b=9\left(1-\frac{2}{3}\right)^2$
$\int-\frac{4}{9}e^xdx$
$\left(\frac{2}{3}x-\frac{1}{2}y\right)^3$
$\left(-6\right)^2-\left(-6\right)-6$
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