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: 79



Work Step by Step

We have $$\int_0^3 \frac{xdx}{x^2+9}=\frac{1}{2}\int_0^3\frac{2xdx}{x^2+9}=\frac{1}{2}\ln (x^2+9)|_0^3\\ =\frac{1}{2}\ln (x^2+9)|_0^3=\frac{1}{2}(\ln 18-\ln 9)= \frac{1}{2}\ln \frac{18}{9}=\frac{1}{2}\ln2.$$
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