Answer
$$
\int \frac{ td t}{t^2+4}=\frac{1}{2}\ln |t^2+4|+c.
$$
Work Step by Step
Le $ u=t^2+4$ and hence $ du=2tdt $, then we have
$$
\int \frac{ td t}{t^2+4}=\frac{1}{2}\int \frac{ d u}{u}= \frac{1}{2}\ln |u|+c=\frac{1}{2}\ln |t^2+4|+c.
$$
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 - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 89
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