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: 91


$$\frac{1}{\sqrt 2} \tan^{-1}(4\sqrt 2).$$

Work Step by Step

We have $$\int_0^4\frac{dx}{2x^2+1}=\frac{1}{\sqrt 2}\int_0^4\frac{d(\sqrt 2x)}{(\sqrt{2}x)^2+1}\\ =\frac{1}{\sqrt 2} \tan^{-1}(\sqrt 2x)|_0^4=\frac{1}{\sqrt 2} \tan^{-1}(4\sqrt 2).$$
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