University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 28


$\ln |\dfrac{\sqrt {e^s+1}-1}{\sqrt {e^s+1}+1}|+c$

Work Step by Step

Consider the integral $\int \dfrac{ds}{\sqrt {e^s+1}}$ Let us take the help of the substitution method. $\sqrt {e^s+1}=a \implies da=e^s ds; ds=\dfrac{(2a) da}{a^2-1}$ The given integral can be re-written as: $\int \dfrac{2da}{a^2-1}=\int \dfrac{(a+1)-(a-1)}{(a-1)(a+1)}da$ This implies that $\int \dfrac{1}{a-1} da- \int \dfrac{1}{a+1} ]da=\ln | \dfrac{a-1}{a+1}|+C$ or, $=\ln |\dfrac{\sqrt {e^s+1}-1}{\sqrt {e^s+1}+1}|+c$
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