Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises: 10

Answer

\[ = \frac{{2x}}{3}{e^{3x}} - \frac{2}{9}{e^{3x}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {2x{e^{3x}}\,\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,then\,\,\,\,du = 2dx \hfill \\ dv = {e^{3x}}dx\,\,\,\,\,\,then\,\,\,\,\,v = \frac{1}{3}{e^{3x}} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ we{\text{ }}replace{\text{ }}the{\text{ }}values{\text{ }}in{\text{ }}the{\text{ }}equation \hfill \\ \hfill \\ \frac{{2x}}{3}{e^{3x}} - \frac{2}{3}\int_{}^{} {{e^{3x}}dx} \hfill \\ \hfill \\ or \hfill \\ \hfill \\ \frac{{2x}}{3}{e^{3x}} - \frac{2}{{3\left( 3 \right)}}\int_{}^{} {{e^{3x}}\left( 3 \right)dx} \hfill \\ \hfill \\ integrate \hfill \\ = \frac{{2x}}{3}{e^{3x}} - \frac{2}{9}{e^{3x}} + C \hfill \\ \end{gathered} \]
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