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: 33

Answer

\[ = \frac{3}{2}\,{\left( {x - 4} \right)^{\frac{3}{2}}} + 8\,{\left( {x - 4} \right)^{\frac{1}{2}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{x}{{\sqrt {x - 4} }}\,dx} \hfill \\ \hfill \\ set \hfill \\ \hfill \\ u = x - 4\,\,\,then\,\,\,s = u + 4 \hfill \\ \hfill \\ and\,\,\,dx = du \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{{u + 4}}{{{u^{\frac{1}{2}}}}}du} \hfill \\ \hfill \\ Distribute \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {{u^{\frac{1}{2}}} + 4{u^{\frac{{ - 1}}{2}}}} \right)du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \frac{3}{2}{u^{\frac{3}{2}}} + 8{u^{\frac{1}{2}}} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = x + 4 \hfill \\ \hfill \\ = \frac{3}{2}\,{\left( {x - 4} \right)^{\frac{3}{2}}} + 8\,{\left( {x - 4} \right)^{\frac{1}{2}}} + C \hfill \\ \hfill \\ \end{gathered} \]
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