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: 22

Answer

\[ = \frac{{\ln \left| {10x - 3} \right|}}{{10}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{1}{{10x - 3}}} \,dx \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = 10x - 3\,\,\,then\,\,\,du = 10xdx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{1}{{10x - 3}}\,dx} = \frac{1}{{10}} \cdot \int_{}^{} {\frac{{10}}{{10x - 3}}} \,dx \hfill \\ \hfill \\ \frac{1}{{10}} \cdot \int_{}^{} {\frac{1}{u}\,du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \frac{{\ln \left| u \right|}}{{10}} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = 10x - 3 \hfill \\ \hfill \\ = \frac{{\ln \left| {10x - 3} \right|}}{{10}} + C \hfill \\ \end{gathered} \]
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