Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 32



Work Step by Step

As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, we have $\int x^{-3}(x+1) dx=\int (x^{-2}+x^{-3})dx$ or, $=-x^{-1}-\dfrac{x^{-2}}{2}+C$ or, $=\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
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