integrate 2x(x^3+4)^3  

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First g(x) = 2x^10 + 24x^7 + 96x^4 + 128x the hard way. Then the integral assuming dx is present is (2/11)x^11 + 3x^8 + (96/5)x^5 + 64x^2 + C Then g'(x) = .2x^9 + (24/7)x^6 + 24x^3 + 128.