Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 28

Answer

\[x + 2\ln \,\left( {x - 2} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{x}{{x - 2}}dx} \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = x - 2\,\,\,\,\,\,then\,\,\,\,\,du = dx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{x - 2}}dx} = \int_{}^{} {\frac{{u + 2}}{u}du} \hfill \\ \hfill \\ = \int_{}^{} {1 + \frac{2}{u}du} \hfill \\ \int_{}^{} {1dx} \,\left( {since\,\,du = dx} \right)\,\,2\int_{}^{} {\frac{1}{u}du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ x + 2\ln \left| u \right| + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,u = x - 2 \hfill \\ \hfill \\ x + 2\ln \,\left( {x - 2} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
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