Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 21

Answer

$G(s)= tan^{-1}+c$

Work Step by Step

$G(s)=\frac{1}{s^2 +1}$ $\int G(s)ds = \int \frac{1}{s^2 +1}$ $G(s)= tan^{-1}+c$
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