Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 53

Answer

$$\frac{1}{5}{\sec ^{ - 1}}\left| {\frac{x}{5}} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{1}{{x\sqrt {{x^2} - 25} }}dx} \cr & {\text{from the table 4}}{\text{.10 }}\int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }}} = \frac{1}{a}{\sec ^{ - 1}}\left| {\frac{x}{a}} \right| + C \cr & {\text{letting }}a = 5,{\text{ }} \cr & = \frac{1}{5}{\sec ^{ - 1}}\left| {\frac{x}{5}} \right| + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{1}{5}{{\sec }^{ - 1}}\left| {\frac{x}{5}} \right| + C} \right) \cr & {\text{ = }}\frac{1}{5}\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}\left| {\frac{x}{5}} \right|} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}\frac{1}{5}\left( {\frac{5}{{x\sqrt {{x^2} - {{\left( 5 \right)}^2}} }}} \right) \cr & {\text{ = }}\frac{1}{{x\sqrt {{x^2} - 25} }} \cr} $$
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