Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 448: 9



Work Step by Step

Multiplying numerator and denominator with $e^{z}$, we get $\int \frac{dz}{e^{z}+e^{-z}}=\int \frac{e^{z}}{e^{2z}+1}dz$ Now, substitute $u=e^{z}$ so that $dz=\frac{du}{e^{z}}$. Then, $\int \frac{e^{z}}{e^{2z}+1}dz=\int \frac{e^{z}}{u^{2}+1}\frac{du}{e^{z}}=\int\frac{1}{u^{2}+1}du$ $=\tan^{-1}u+C=\tan^{-1}(e^{z})+C$
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