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: 65


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

Work Step by Step

Since $ u=\ln t $, then $ du=\frac{1}{t}dt $. Now, we have $$\int \frac{dt}{t(1+(\ln t)^2)}dt =\int \frac{1}{1+u^2}du\\ =\tan^{-1}u+c=\tan^{-1}(\ln t)+c.$$
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