Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 54

Answer

$${\cosh ^{ - 1}}\frac{x}{4} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {{x^2} - 16} }}} \cr & {\text{rewriting the denominator}} \cr & \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( 4 \right)}^2}} }}} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, formula 1}} \cr & \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = {\cosh ^{ - 1}}\frac{x}{a} + C \cr & ,so \cr & = {\cosh ^{ - 1}}\frac{x}{4} + C \cr} $$
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