Answer
$ln|sec(x)-1|+C$
Work Step by Step
$\int \frac{sec(x)+tan(x)}{sec(x)-1}dx$
let $u=sec(x)-1$
$du=sec(x)+tan(x)dx$
$\int \frac{sec(x)+tan(x)}{sec(x)-1}dx$
$=\int \frac{1}{u}du$
$=ln|sec(x)-1|+C$
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 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 39
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