Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 22


$\int x(x^2-4)dx = \frac{1}{4}x^4 -2x^2+C$

Work Step by Step

$\int x(x^2-4)dx = \int(x^3-4x)dx = \frac{1}{4}x^4 -4\times\frac{1}{2}x^2 +C = \frac{1}{4}x^4 -2x^2+C$
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