How do you integrate #int (x^22x2)^2 dx# ?

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Answer 1 · 2017-04-02T10:34:20

#16x^7+c#

First let's simplify the integrand.

#(x^2*2x*2)^2 = (4x^3)^2 = 16x^6#

#I= int 16x^6 dx #

#= 16 * int x^6 dx#

#= 16 * x^7/7 +c# [Power rule}

#=16x^7+c#

How do you integrate #int (x^2*2x*2)^2 dx# ?