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 - Page 515: 47

Answer

\[ = 2{\tan ^{ - 1}}\sqrt x + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{{x^{\frac{1}{2}}} + {x^{\frac{3}{2}}}}}} \hfill \\ \hfill \\ {\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{dx}}{{{x^{\frac{1}{2}}}\,\left( {1 + x} \right)}}} \hfill \\ \hfill \\ substituting\,\,\,x = {t^2}\,\, \to \,dx\, = 2tdt \hfill \\ \hfill \\ = \int_{}^{} {\frac{{2tdt}}{{t + {t^3}}}} = \,\,\,\,2\int_{}^{} {\frac{{t\,dt}}{{t\,\left( {1 + {t^2}} \right)}}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = 2\int_{}^{} {\frac{{dt}}{{1 + {t^2}}}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = 2{\tan ^{ - 1}}t + C \hfill \\ \hfill \\ substituting\,back\,\,t = \,\sqrt x \hfill \\ \hfill \\ = 2{\tan ^{ - 1}}\sqrt x + C \hfill \\ \hfill \\ \end{gathered} \]
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