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: 35

Answer

\[ = \frac{3}{5}\,{\left( {x + 4} \right)^{\frac{5}{3}}} - 6\,{\left( {x + 4} \right)^{\frac{2}{3}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{x}{{\sqrt[3]{{x + 4}}}}\,dx} \hfill \\ \hfill \\ Set\,\,\,u = x + 4\,\,,\,\,so \hfill \\ \hfill \\ x = u - 4\,\,\,\,\,{\text{which implies that}}\,\,\,\,dx = du\,\, \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{{u - 4}}{{{u^{\frac{1}{3}}}}}\,du} \hfill \\ \hfill \\ distribute \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {{u^{\frac{2}{3}}} - 4{u^{ - \frac{1}{3}}}} \right)} \,du \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{3}{5}{u^{\frac{5}{3}}} - 6{u^{\frac{2}{3}}} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = x + 4 \hfill \\ \hfill \\ = \frac{3}{5}\,{\left( {x + 4} \right)^{\frac{5}{3}}} - 6\,{\left( {x + 4} \right)^{\frac{2}{3}}} + C \hfill \\ \hfill \\ \end{gathered} \]
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