Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises: 25

Answer

\[\frac{2}{3}{\left( {u - 1} \right)^{3/2}} + 2\,{\left( {u - 1} \right)^{1/2}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{u}{{\sqrt {u - 1} }}du} \hfill \\ Let\,\,z = u - 1\,\,,\,\,Then\,\,u = z + 1 \hfill \\ So\,\,that\,\,\,du = dz \hfill \\ Then \hfill \\ \int_{}^{} {\frac{u}{{\sqrt {u - 1} }}du} = \int_{}^{} {\frac{{z + 1}}{{\sqrt z }}dz} \hfill \\ Write\,\,the\,\,integrand\,\,as \hfill \\ \int_{}^{} {\,\left( {\frac{z}{{\sqrt z }} + \frac{1}{{\sqrt z }}} \right)dz} = \int_{}^{} {\,\left( {{z^{1/2}} + {z^{ - 1/2}}} \right)dz} \hfill \\ Use\,\,power\,\,rule \hfill \\ \frac{{{z^{3/2}}}}{{3/2}} + \frac{{{z^{1/2}}}}{{1/2}} + C \hfill \\ \frac{2}{3}{z^{3/2}} + 2{z^{1/2}} + C \hfill \\ Substituting\,\,u = u - 1\,\,gives \hfill \\ \frac{2}{3}{\left( {u - 1} \right)^{3/2}} + 2\,{\left( {u - 1} \right)^{1/2}} + C \hfill \\ \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.