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: 16

Answer

\[ = \frac{1}{{120}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{225 - 16{x^3}}}} \hfill \\ \hfill \\ {\text{rewrite}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ = \frac{1}{4}\int_{}^{} {\frac{{4dx}}{{{{15}^2} - \,{{\left( {4x} \right)}^2}}}} \hfill \\ \hfill \\ use\,\,\int_{}^{} {\frac{{du}}{{{a^2} - {u^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \hfill \\ \hfill \\ with\,\,a = 15{\text{ and }}u = 4x \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{1}{{\,\left( 2 \right)\,\left( {15} \right)\,\left( 4 \right)}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{1}{{120}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C \hfill \\ \end{gathered} \]
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.