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: 11

Answer

\[ = \frac{{\,{{\left( {\ln 2x} \right)}^2}}}{2} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\ln 2x}}{x}\,dx} \hfill \\ \hfill \\ set\,\,\,\,\ln \,2x = \,u\,\,\,and\,\,\,\,\frac{1}{{2x}}2\,dx = du\,\,or\,\frac{{dx}}{x} = du \hfill \\ \hfill \\ \operatorname{int} egral\,becomes \hfill \\ \hfill \\ \int_{}^{} {\frac{{\ln 2x}}{x}\,dx} = \int_{}^{} {u\,du} \hfill \\ \hfill \\ {\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = \frac{{{u^2}}}{2} + C \hfill \\ \hfill \\ substituting\,back\,\,\,u = \ln 2x \hfill \\ \hfill \\ = \frac{{\,{{\left( {\ln 2x} \right)}^2}}}{2} + C \hfill \\ \end{gathered} \]
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