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 425: 23

Answer

$\frac{2}{3}\sqrt {3ax + b{x^3}} + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{a + b{x^2}}}{{\sqrt {3ax + b{x^3}} }}} dx \cr & {\text{Let }}u = 3ax + b{x^3},{\text{ then }}du = \left( {3a + 3b{x^2}} \right)dx \cr & du = 3\left( {a + b{x^2}} \right)dx \cr & \frac{1}{3}du = \left( {a + b{x^2}} \right)dx \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & \int {\frac{{a + b{x^2}}}{{\sqrt {3ax + b{x^3}} }}} dx = \int {\frac{1}{{\sqrt u }}\left( {\frac{1}{3}} \right)} du \cr & = \frac{1}{3}\int {\frac{1}{{\sqrt u }}} du \cr & = \frac{1}{3}\int {{u^{ - 1/2}}} du \cr & {\text{Integrate}} \cr & = \frac{1}{3}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr & = \frac{2}{3}{u^{1/2}} + C \cr & = \frac{2}{3}\sqrt u + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}u = 3ax + b{x^3} \cr & = \frac{2}{3}\sqrt {3ax + b{x^3}} + C \cr} $$
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