Calculus (3rd Edition)

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


$$\frac{1}{2} \tan^{-1}2x+c.$$

Work Step by Step

Since $u=2x$, then $du=2dx$, and hence we get $$\int \frac{ dx}{4x^2+1}=\frac{1}{2}\int \frac{ du}{ u^2+1} =\frac{1}{2} \tan^{-1}u+c\\ =\frac{1}{2} \tan^{-1}2x+c.$$
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