$\frac{8x^4+12x^3+6x^2+x}{3}$
$\int\left(4x^6-2x^3+7x-4\right)dx$
$\left(-a\:\right)^4$
$16\:x\:\frac{3}{4}$
$5\:cos\:x\:=\:0$
$\int\frac{1}{3}\left(x+a\right)dx$
$\:-4\left(x\:-\:3y\right)\:-\:2\left(8x\:+\:4y\right)\:$
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