University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 32


$\dfrac{1}{8} \ln (25+4x^2) +C$

Work Step by Step

Let us consider $u=25 +4x^2 \implies du =(8x) dx$ Now, the given integral can be written as: $\dfrac{1}{8}\int\dfrac{1}{u} du=\dfrac{1}{8} \ln |u|+C $ or, $=\dfrac{1}{8} \ln (25+4x^2) +C$
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