Answer
$$\frac{\pi}{12}.$$
Work Step by Step
We have
$$\int_0^3 \frac{dx}{x^2+9}=\frac{1}{3}\int_0^3\frac{d(x/3)}{(x/3)^2+1}=\frac{1}{3}\tan^{-1} (x/3)|_0^3\\
=\frac{1}{3}(\tan^{-1} 1-\tan^{-1}0)=\frac{\pi}{12}.$$
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 - Chapter Review Exercises - Page 387: 80
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
