Answer
$G(s)= tan^{-1}+c$
Work Step by Step
$G(s)=\frac{1}{s^2 +1}$
$\int G(s)ds = \int \frac{1}{s^2 +1}$
$G(s)= tan^{-1}+c$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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 - Page 327: 21
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