Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 579: 12

Answer

$$\frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{4}{{15}}{\left( {x - 1} \right)^{5/2}} + C$$

Work Step by Step

$$\eqalign{ & \int {x\sqrt {x - 1} } dx \cr & {\text{Integrating by parts}} \cr & u = x,{\text{ }}du = dx \cr & dv = \sqrt {x - 1} dx,{\text{ }}v = \int {{{\left( {x - 1} \right)}^{1/2}}dx} ,{\text{ }}v = \frac{2}{3}{\left( {x - 1} \right)^{3/2}} \cr & \int {udv} = uv - \int {vdu} \cr & \int {x\sqrt {x - 1} } dx = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \int {\frac{2}{3}{{\left( {x - 1} \right)}^{3/2}}dx} \cr & {\text{Integrating}} \cr & = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{2}{3}\int {{{\left( {x - 1} \right)}^{3/2}}dx} \cr & = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{2}{3}\left( {\frac{{{{\left( {x - 1} \right)}^{5/2}}}}{{5/2}}} \right) + C \cr & = \frac{2}{3}x{\left( {x - 1} \right)^{3/2}} - \frac{4}{{15}}{\left( {x - 1} \right)^{5/2}} + C \cr} $$
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