$\frac{d}{dx}\left(\left(4x+1\right)\left(2x^2+x-3\right)^4\right)=4\left(2x^2+x-3\right)^4+4\left(4x+1\right)\left(2x^2+x-3\right)^{3}\left(4x+1\right)$
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Integral
$\int\left(4x+1\right)\left(2x^2+x-3\right)^4dx=\frac{\left(2x^2+x-3\right)^{5}}{5}+C_0$
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