Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 57


See the proof below.

Work Step by Step

Since $u=x/3$, then $du= \frac{1}{3}dx$, and hence we get $$\int \frac{ dx}{9+x^2}=\int \frac{3 du}{9+9u^2}=\frac{1}{3}\int \frac{ du}{1+u^2}\\ =\frac{1}{3} \tan^{-1}u+c=\frac{1}{3} \tan^{-1}\frac{x}{3}+c.$$
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