Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 39



Work Step by Step

$\int \frac{sec(x)+tan(x)}{sec(x)-1}dx$ let $u=sec(x)-1$ $du=sec(x)+tan(x)dx$ $\int \frac{sec(x)+tan(x)}{sec(x)-1}dx$ $=\int \frac{1}{u}du$ $=ln|sec(x)-1|+C$
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