Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 79

Answer

\[ = \frac{1}{{{a^2}}}\,\,\,\,\left[ {\,\left( {ax + b} \right) - b\ln \left| {ax + b} \right|} \right] + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{x}{{ax + b}}\,\,dx} \hfill \\ \hfill \\ set = ax + b\,\,\,then\,\,\,x = \frac{{u - b}}{a}{\text{ }}and\,\,du = adx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{ax + b}}\,\,dx} = \frac{1}{a}\int_{}^{} {\,\left( {\frac{{u - b}}{{au}}} \right)du} \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{ax + b}}\,\,dx} = \frac{1}{{{a^2}}}\int_{}^{} {\,\frac{{u - b}}{u}du} \hfill \\ \hfill \\ = \frac{1}{{{a^2}}}\int_{}^{} {\,\left( {1 - \frac{b}{u}} \right)du} \hfill \\ \hfill \\ = \frac{1}{{{a^2}}}\,\,\,\,\left[ {u - b\ln \left| u \right|} \right] + C \hfill \\ \hfill \\ substitute\,\,back \hfill \\ \hfill \\ = \frac{1}{{{a^2}}}\,\,\,\,\left[ {\,\left( {ax + b} \right) - b\ln \left| {ax + b} \right|} \right] + C \hfill \\ \end{gathered} \]
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