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Posted by 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

  • Calculus - , Sunday, August 5, 2012 at 6:36am


    then du = 2(3x^2) dx

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