Answer
$$\frac{1}{2} \tan^{-1}2x+c.$$
Work Step by Step
Since $u=2x$, then $du=2dx$, and hence we get
$$\int \frac{ dx}{4x^2+1}=\frac{1}{2}\int \frac{ du}{ u^2+1}
=\frac{1}{2} \tan^{-1}u+c\\
=\frac{1}{2} \tan^{-1}2x+c.$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 58
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