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: 80

Answer

\[ = \frac{{2\sqrt {ax + b} }}{{{a^2}}}\,\,\left[ {\,\frac{{\,\left( {ax - 2b} \right)}}{3}} \right] + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} \hfill \\ \hfill \\ set\,\,\,{u^2} = ax + b\,\,\,\,then\,\,\,\,x = \frac{{{u^2} - b}}{a} \hfill \\ and \hfill \\ 2udu = adx\,\,\, \to dx = \frac{{2udu}}{a} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{1}{a}\int_{}^{} {\,\left( {\frac{{{u^2} - b}}{{au}}} \right)\,\left( {2u} \right)du} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{2}{{{a^2}}}\int_{}^{} {\,\left( {{u^2} - b} \right)} du \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{2}{{{a^2}}}\,\,\left[ {\frac{{{u^3}}}{3} - bu} \right] + C \hfill \\ substitute\,\,back \hfill \\ \hfill \\ = \frac{{2\sqrt {ax + b} }}{{{a^2}}}\,\,\left[ {\,\frac{{\,\left( {ax - 2b} \right)}}{3}} \right] + C \hfill \\ \end{gathered} \]
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