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: 80



Work Step by Step

We have $$\int_0^3 \frac{dx}{x^2+9}=\frac{1}{3}\int_0^3\frac{d(x/3)}{(x/3)^2+1}=\frac{1}{3}\tan^{-1} (x/3)|_0^3\\ =\frac{1}{3}(\tan^{-1} 1-\tan^{-1}0)=\frac{\pi}{12}.$$
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