What is #int (x^3+8) *(x^2-2x+4)dx#?

1 Answer
Trevor Ryan.
Nov 1, 2015

#1/6x^6-2/5x^5+x^4+8/3x^3-8x^2+32x+C#

Explanation:

The best would just be to multiply the whole thing out and then integrate term by term.

#int(x^3+8)(x^2-2x+4)dx=int(x^5-2x^4+4x^3+8x^2-16x+32)dx#

#=x^6/6-2x^5/5+4x^4/4+8x^3/3-16x^2/2+32x+C#

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