$\int\left(8x^2\left(4x^2+12\right)^{-3}\right)dx$
$0+120+-4$
$\left(2x-3x+1\right)+\left(4x^2-6x-8\right)$
$0.00139\cdot10^2$
$\frac{d}{dx}\left(\left(2x-7\right)^3\sin\left(x\right)^{\cos\left(x\right)}\right)$
$8+9+7+10+7+9$
$y'+\left(\frac{1}{x}\right)y=x$
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