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: 23

Answer

\[ = \frac{{\,{{\left( {{x^4} + 16} \right)}^7}}}{{28}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{x^3}\,{{\left( {{x^4} + 16} \right)}^6}} \,dx \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = {x^4} + 16\,\,\,\,\,\,then\,\,\,\,du = 4{x^3}dx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {{x^3}\,{{\left( {{x^4} + 16} \right)}^6}} \,dx = \frac{1}{4} \cdot \int_{}^{} {4{x^3}\,{{\left( {{x^4} + 16} \right)}^6}dx} \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ = \frac{1}{4} \cdot \int_{}^{} {{u^6}\,du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \frac{1}{4} \cdot \frac{{{u^7}}}{7} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,u = {x^4} + 16 \hfill \\ \hfill \\ = \frac{{\,{{\left( {{x^4} + 16} \right)}^7}}}{{28}} + C \hfill \\ \end{gathered} \]
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