Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 90



Work Step by Step

Using the method of partial fractions, we have $$\int_0^1\frac{dx}{25-x^2}=\frac{1}{10}\int_0^1 \frac{1}{5-x}+\frac{1}{5+x}dx \\ =\frac{1}{10}(-\ln(5-x)+\ln(5+x))|_0^1=\frac{1}{10}(-\ln4+\ln6+\ln5-\ln5)\\ =\frac{1}{10} \ln \frac{6}{4}=\frac{1}{10}\ln\frac{3}{2}=0.0405.$$
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