Answer
$\tan{y}$
Work Step by Step
$\int(\tan^2{y}+1)dy=\int\sec^2{y}dy=\tan{y}+C$
CHECK:
$\frac{d}{dy}(\tan{y}+C)=\frac{d}{dy}\tan{y}+\frac{d}{dy}C=\sec^2{y}=0=\sec^2{y}=1+\tan^2{y}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.1 Exercises - Page 251: 31
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