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∫(x+4)(x−4)dx=x33−16x+C\int(x+4)(x-4) d x=\frac{x^{3}}{3}-16 x+C

Question 149

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∫(x+4)(x−4)dx=x33−16x+C\int(x+4)(x-4) d x=\frac{x^{3}}{3}-16 x+C .

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