Answer
$\tan^{-1}(\ln t)+c.$
Work Step by Step
Since $ u=\ln t $, then $ du=\frac{1}{t}dt $.
Now, we have
$$\int \frac{dt}{t(1+(\ln t)^2)}dt =\int \frac{1}{1+u^2}du\\
=\tan^{-1}u+c=\tan^{-1}(\ln t)+c.$$
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: 65
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