Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 515: 42

Answer

$$ - 2 - \ln 4$$

Work Step by Step

$$\eqalign{ & \int_4^9 {\frac{{dx}}{{1 - \sqrt x }}} \cr & {\text{Multiply the integrand by }}\frac{{2\sqrt x }}{{2\sqrt x }} \cr & = \int {\frac{1}{{1 - \sqrt x }}\left( {\frac{{2\sqrt x }}{{2\sqrt x }}} \right)} dx \cr & = \int {\frac{{2\sqrt x }}{{2\sqrt x \left( {1 - \sqrt x } \right)}}} dx \cr & \cr & {\text{Let }}u = \sqrt x ,\,\,\,du = \frac{1}{{2\sqrt x }}dx \cr & {\text{Use the substitution}} \cr & = \int {\frac{{2u}}{{1 - u}}} du \cr & = - 2\int {\frac{u}{{u - 1}}} du \cr & = - 2\int {\left( {1 + \frac{1}{{u - 1}}} \right)} du \cr & = - 2\int {du} - 2\int {\frac{1}{{u - 1}}} du \cr & \cr & {\text{Integrating}} \cr & {\text{ = }} - 2u - 2\ln \left| {u - 1} \right| + C \cr & {\text{substitute }}u = \sqrt x \cr & {\text{ = }} - 2\sqrt x - 2\ln \left| {\sqrt x - 1} \right| + C \cr & {\text{,then}} \cr & \int_4^9 {\frac{{dx}}{{1 - \sqrt x }}} = \left[ { - 2\sqrt x - 2\ln \left| {\sqrt x - 1} \right|} \right]_4^9 \cr & = \left( { - 2\sqrt 9 - 2\ln \left| {\sqrt 9 - 1} \right|} \right) - \left( { - 2\sqrt 4 - 2\ln \left| {\sqrt 4 - 1} \right|} \right) \cr & = \left( { - 6 - 2\ln \left| 2 \right|} \right) - \left( { - 4 - 2\ln \left| 1 \right|} \right) \cr & = - 6 - \ln 4 + 4 \cr & = - 2 - \ln 4 \cr} $$
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