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: 20

Answer

\[ = \frac{{{{\left( {\sqrt x + 1\,} \right)}^5}}}{5} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\,{{\left( {\sqrt x + 1} \right)}^4}}}{{2\sqrt x }}} \,dx \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = \sqrt x + 1\,\,\,\,\,\,then\,\,\,\,\,du = \frac{1}{{2\sqrt x }}dx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{\,{{\left( {\sqrt x + 1} \right)}^4}}}{{2\sqrt x }}} \,dx = \int_{}^{} {\,{{\left( {\sqrt x + 1} \right)}^4} \cdot \frac{1}{{2\sqrt x }}\,dx} \hfill \\ \hfill \\ \int_{}^{} {{u^4}du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ \frac{{{u^5}}}{5} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,u = \sqrt x + 1 \hfill \\ \hfill \\ = \frac{{{{\left( {\sqrt x + 1\,} \right)}^5}}}{5} + C \hfill \\ \end{gathered} \]
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