Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 17

Answer

\[ = \frac{{\,{{\left( {{x^2} - 1} \right)}^{100}}}}{{100}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {2x\,{{\left( {{x^2} - 1} \right)}^{99}}} \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = {x^2} - 1\,\,\,\,then\,\,\,du = 2xdx \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \int_{}^{} {2x\,{{\left( {{x^2} - 1} \right)}^{99}}\,\,} \,\, = \int_{}^{} {{u^{99}}\,du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \frac{{{u^{100}}}}{{100}} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,u = {x^2} - 1 \hfill \\ \hfill \\ = \frac{{\,{{\left( {{x^2} - 1} \right)}^{100}}}}{{100}} + C \hfill \\ \end{gathered} \]
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