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: 53

Answer

$$\frac{1}{{2\sqrt 2 }}{\coth ^{ - 1}}\frac{x}{{2\sqrt 2 }} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{8 - {x^2}}}} , \cr & {\text{rewriting the denominator}} \cr & \int {\frac{{dx}}{{{{\left( {\sqrt 8 } \right)}^2} - {x^2}}}} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, with }}x{\text{ > 2}}\sqrt 2 \cr & \int {\frac{{dx}}{{{{\left( {\sqrt 8 } \right)}^2} - {x^2}}} = \frac{1}{{\sqrt 8 }}{{\coth }^{ - 1}}\frac{x}{{\sqrt 8 }} + C} \cr & \sqrt 8 = 2\sqrt 2 ,then \cr & = \frac{1}{{2\sqrt 2 }}{\coth ^{ - 1}}\frac{x}{{2\sqrt 2 }} + C \cr} $$
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