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 - Page 503: 59

Answer

$$\ln \left( {2 + \sqrt 5 } \right)$$

Work Step by Step

$$\eqalign{ & \int_1^{{e^2}} {\frac{{dx}}{{x\sqrt {{{\ln }^2}x + 1} }}} \cr & {\text{set }}u = \ln x{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \frac{1}{x}dx, \cr & {\text{switch the limits of integration}} \cr & u = \ln x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \to u = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = {e^2} \to u = 2 \cr & {\text{use the change of variable}} \cr & \int_1^{{e^2}} {\frac{{dx}}{{x\sqrt {{{\ln }^2}x + 1} }}} = \int_0^2 {\frac{{du}}{{\sqrt {{u^2} + 1} }}} \cr & {\text{using the Integral formulas in the Theorem 6}}{\text{.12 }}\left( {formula\,\,2} \right)\,\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = {{\sinh }^{ - 1}}\frac{x}{a} + C{\text{.}}} \cr & {\text{then}} \cr & \int_0^2 {\frac{{du}}{{\sqrt {{u^2} + 1} }}} = \left( {{{\sinh }^{ - 1}}u} \right)_0^2 \cr & {\text{evaluate the limits of integration}} \cr & = {\sinh ^{ - 1}}2 - {\sinh ^{ - 1}}0 \cr & = {\sinh ^{ - 1}}2 \cr & {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}: \cr & {\text{ }}{\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right){\text{ then}}{\text{,}} \cr & {\sinh ^{ - 1}}2 = \ln \left( {2 + \sqrt {{2^2} + 1} } \right) \cr & = \ln \left( {2 + \sqrt 5 } \right) \cr} $$
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