Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 426: 28

Answer

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

Work Step by Step

$$\eqalign{ & \int {x\sqrt {x + 2} } dx \cr & {\text{Let }}u = x + 2 \to x = u - 2,{\text{ }}dx = du \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & \int {x\sqrt {x + 2} } dx = \int {\left( {u - 2} \right)\sqrt u } du \cr & = \int {\left( {u - 2} \right){u^{1/2}}} du \cr & = \int {\left( {{u^{3/2}} - 2{u^{1/2}}} \right)} du \cr & {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr & = \frac{{{u^{5/2}}}}{{5/2}} - 2\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr & = \frac{{2{u^{5/2}}}}{5} - \frac{{4{u^{3/2}}}}{3} + C \cr & {\text{Write in terms of }}t,{\text{ substitute }}u = x + 2 \cr & = \frac{{2{{\left( {x + 2} \right)}^{5/2}}}}{5} - \frac{{4{{\left( {x + 2} \right)}^{3/2}}}}{3} + C \cr} $$
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