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: 54

Answer

\[ = \frac{8}{9}\]

Work Step by Step

\[\begin{gathered} \int_0^2 {\frac{2}{{{s^3} + 3{s^2} + 3s + 1}}\,ds} \hfill \\ \hfill \\ factor\,ing\,\,the\,\,denominator \hfill \\ \hfill \\ = \int_0^2 {\frac{2}{{\,{{\left( {s + 1} \right)}^3}}}\,ds} \hfill \\ \hfill \\ set\,\,s + 1\, = \,t\,\,\,\, \to \,\,\,ds = dt \hfill \\ \hfill \\ = \int_1^3 {\frac{2}{{{t^3}}}} dt\,\, \hfill \\ \hfill \\ \,integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \frac{{ - 2}}{2}\,\,\left[ {\frac{1}{{{t^2}}}} \right]_1^3 \hfill \\ \hfill \\ = - \,\left( {\frac{1}{9} - 1} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = \frac{8}{9} \hfill \\ \end{gathered} \]
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