Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 31

Answer

$$2{\sec ^{ - 1}}2x + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{2}{{x\sqrt {4{x^2} - 1} }}dx} \cr & {\text{substitute }}u = 2x,{\text{ }}du = 2dx{\text{ and }}dx = \frac{{du}}{2} \cr & \int {\frac{2}{{x\sqrt {4{x^2} - 1} }}dx} = \int {\frac{2}{{\left( {u/2} \right)\sqrt {{u^2} - 1} }}} \frac{{du}}{2} \cr & = 2\int {\frac{1}{{u\sqrt {{u^2} - 1} }}} du \cr & {\text{find the antiderivative}} \cr & = 2{\sec ^{ - 1}}u + C \cr & {\text{ with}}\,\,\,u = 2x \cr & = 2{\sec ^{ - 1}}2x + C \cr} $$
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