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: 62


$$\int \frac{dx}{4x^2+9}=\frac{1}{6} \tan^{-1 }(\frac{2}{3}x)+c.$$

Work Step by Step

Let $ u=\frac{2}{3}x $, then $ du= \frac{2}{3}dx $ and hence $$\int \frac{dx}{4x^2+9}=\frac{3}{2}\int \frac{du}{9u^2+9} =\frac{3}{2} \frac{1}{9}\int \frac{du}{u^2+1}\\ =\frac{1}{6} \tan^{-1 }u+c=\frac{1}{6} \tan^{-1 }(\frac{2}{3}x)+c.$$
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