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: 7

Answer

\[ = \frac{1}{{15\,{{\left( {3 - 5x} \right)}^3}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{\,{{\left( {3 - 5x} \right)}^4}}}} \hfill \\ \hfill \\ substituting\, \hfill \\ \hfill \\ 3 - 5x = u\,\,\,\,and\,\,\, - 5x = du\,\,or\,\,\,dx = - \frac{{du}}{5} \hfill \\ \hfill \\ - \frac{1}{5}\int_{}^{} {\frac{{du}}{{{u^4}}}} = - \frac{1}{5}\int {{u^{ - 4}}du} \hfill \\ \hfill \\ {\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = - \frac{1}{5}\,\left( { - \frac{1}{{3{u^3}}}} \right) + C = \frac{1}{{15{u^3}}} + C \hfill \\ \hfill \\ substituting\,back\,u = \,\left( {3 - 5x} \right) \hfill \\ \hfill \\ = \frac{1}{{15\,{{\left( {3 - 5x} \right)}^3}}} + C \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.