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: 48

Answer

\[ = - 2\,\,\left[ {1 + 4\ln \,\left( {\frac{3}{4}} \right)} \right]\]

Work Step by Step

\[\begin{gathered} \int_0^1 {\frac{{dp}}{{4 - \sqrt p }}} \hfill \\ \hfill \\ substituting\,\,p = {t^2}\, \to \,\,dp = 2tdt \hfill \\ \hfill \\ = \int_0^1 {\frac{{2tdt}}{{4 - t}}} \hfill \\ \hfill \\ pull\,\,out\,\,the\,\,{\text{constant}} \hfill \\ \hfill \\ = 2\int_0^1 {\frac{{tdt}}{{4 - t}}} \hfill \\ \hfill \\ {\text{using}}\,\,{\text{the long division}} \hfill \\ \hfill \\ = - 2\int_0^1 {\,\left( {1 + \frac{4}{{t - 4}}} \right)dt} \hfill \\ \hfill \\ \,integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = - 2\,\,\left[ {t + 4\ln \left| {t - 4} \right|} \right]_0^1 \hfill \\ \hfill \\ = - 2\,\,\left[ {1 + 4\ln 3 - 0 - 4\ln 4} \right] \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = - 2\,\,\left[ {1 + 4\ln \,\left( {\frac{3}{4}} \right)} \right] \hfill \\ \end{gathered} \]
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