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: 22

Answer

\[\frac{{{y^2}}}{2} - \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dy}}{{{y^{ - 1}} + {y^{ - 3}}}}} \,\, = \int_{}^{} {\frac{{{y^3}dy}}{{{y^2} + 1}}} \hfill \\ \hfill \\ {\text{using}}\,\,\,{\text{the}}\,\,{\text{long}}\,\,\,{\text{division}} \hfill \\ \hfill \\ \int_{}^{} {\,\left( {y - \frac{y}{{{y^2} + 1}}} \right)} dy \hfill \\ \hfill \\ split\,\,the\,\,{\text{integrand}} \hfill \\ \hfill \\ \int_{}^{} {\,y} dy - \int_{}^{} {\,\frac{y}{{{y^2} + 1}}} dy \hfill \\ \hfill \\ or \hfill \\ \hfill \\ \int_{}^{} {\,y} dy - \frac{1}{2}\int_{}^{} {\,\frac{{2y}}{{{y^2} + 1}}} dy \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ \frac{{{y^2}}}{2} - \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C \hfill \\ \end{gathered} \]
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