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Integrate it step by step:

∫((2x+4)(2x-1))/(3x-2)

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Just do a long division first:
((2x+4)(2x-1))/(3x-2) = 4x/3 + 26/9 + 16/9 * 1/(3x-2)
so,
∫((2x+4)(2x-1))/(3x-2) dx = 2x^2/3 + 26/9 x + 16/27 log(3x-2) + C

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