Answer
$7$
Work Step by Step
Using the fact that $\int\frac{dx}{x^2+1}=\tan^{-1}x$, we have
$$\int^{\tan 8}_{\tan 1}\frac{dx}{x^2+1}=\tan^{-1}x|^{\tan 8}_{\tan 1}\\
=\tan^{-1}\tan 8-\tan^{-1}\tan 1=8-1=7.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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