Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 56


$ \frac{1}{2}(\tan ^{-1} t)^2+c.$

Work Step by Step

We have $$ \int \frac{\tan ^{-1} t d t}{1+t^{2}}=\int \tan ^{-1} t \ d (\tan ^{-1} t)\\ = \frac{1}{2}(\tan ^{-1} t)^2+c. $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.