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: 31

Answer

\[ = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + t - 3\ln \,\left| {t + 1} \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{t^3} - 2}}{{t + 1}}dt} \hfill \\ \hfill \\ {\text{using}}\,\,{\text{the}}\,\,{\text{long}}\,\,{\text{division}} \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {{t^2} - t + 1 - \frac{3}{{t + 1}}} \right)dt} \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ = \int_{}^{} {\,{t^2}dt} - \int_{}^{} {\,tdt} + \int_{}^{} {\,dt} - \int_{}^{} {\frac{3}{{t + 1}}\,dt} \hfill \\ \hfill \\ integrate\, \hfill \\ \hfill \\ = \frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} + t - 3\ln \,\left| {t + 1} \right| + C \hfill \\ \end{gathered} \]
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