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March 27, 2015

March 27, 2015

Posted by **renee** on Tuesday, June 28, 2011 at 12:17pm.

(X^4-16)^5 X^3 DX

- CALCULUS -
**MathMate**, Tuesday, June 28, 2011 at 5:05pmHint:

note the x^4 in parentheses, and that d(x^4)/dx = 4x^3, so substituting

y=x^4 would have a good chance of success.

If

y=x^4-16,

dy=4x^3dx

So

I=∫(X^4-16)^5 X^3 DX

=∫y^5(dy/4)

proceed with the power rule and put back y=x^4-16 into I.

Do not forget the integration constant C.

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