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posted by Cynthia on Saturday, August 4, 2012 at 9:44pm.

to evaluate (integral) 3x^2 cos (2x^3-4) dx, it is necessary to let A. u=3x^2 B. u=6x C. u=2x^3-4 D. u=6x^2

C then du = 2(3x^2) dx

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