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: 43


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

Work Step by Step

Consider $x^2=\tan \theta \implies 2x dx= \sec^2 \theta d \theta $ Now, the given integral can be written as: $\dfrac{1}{2}\int\dfrac{\sec^2 \theta}{ \sqrt{1 +\tan^2 \theta}} d\theta= \dfrac{1}{2} \int \sec \theta d\theta$ or, $=\dfrac{1}{2}\ln |\sec d \theta+\tan d \theta|+C$ or, $=\dfrac{1}{2}\ln (x^2+\sqrt{x^4+1})+C$
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