University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 40



Work Step by Step

Given: $\int x^{-3}(x+1) dx$ Since, we have $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, $\int x^{-3}(x+1) dx=\int (x^{-2}+x^{-3})dx=-x^{-1}-\dfrac{x^{-2}}{2}+C=\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
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