Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 14

Answer

\[ = {e^2}^{\sqrt y + 1} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{e^{2\sqrt y + 1}}}}{{\sqrt y }}dy} \hfill \\ \hfill \\ set\,\,\,\left( {2\sqrt y + 1} \right) = u\,\,then\,\,\frac{{dy}}{{\sqrt y }} = du \hfill \\ \hfill \\ integral\,\,becomes \hfill \\ \hfill \\ \int_{}^{} {\frac{{{e^{2\sqrt y + 1}}}}{{\sqrt y }}dy} = \int_{}^{} {{e^u}du} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = {e^u} + C \hfill \\ \hfill \\ substituting\,back\,\,\,u = 2\sqrt y + 1 \hfill \\ \hfill \\ = {e^2}^{\sqrt y + 1} + C \hfill \\ \end{gathered} \]
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