Answer
$\frac{1}{3}tan(3x+2)+C$
Work Step by Step
Let u=3x+2 $=>du=3dx=>\frac{1}{3}du=dx$
=$\int sec^2(3x+2)dx$
=$\int (sec^2u)(\frac{1}{3}du)$
=$\frac{1}{3}\int sec^2udu$
=$\frac{1}{3}tan u+C$
=$\frac{1}{3}tan(3x+2)+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 23
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