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: 53



Work Step by Step

Using the fact that $\int\frac{dx}{x^2+1}=\tan^{-1}x$, we have $$\int^{\tan 8}_{\tan 1}\frac{dx}{x^2+1}=\tan^{-1}x|^{\tan 8}_{\tan 1}\\ =\tan^{-1}\tan 8-\tan^{-1}\tan 1=8-1=7.$$
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