$\int\left(4-2t^2\right)^7\cdot tdt$
$8 x ^ { 3 } + 6 x$
$\int6.dt$
$-\left(-1+4\right)-\left(-7+1\right)$
$\left(n-8\right)\left(n+12\right)$
$\frac{d}{dx}\left(\frac{1}{10}x^5+\frac{1}{6}x^{-3}\right)$
$6x^3+12x+16.$
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