Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 19



Work Step by Step

$$A=\int\frac{a+bx^2}{\sqrt{3ax+bx^3}}dx$$ Let $u=3ax+bx^3$. Then $du=(3a+3bx^2)dx=3(a+bx^2)dx$, so $(a+bx^2)dx=\frac{1}{3}du$ Also, $\sqrt{3ax+bx^3}=\sqrt u=u^{1/2}$ Substitute into $A$, we have $$A=\frac{1}{3}\int\frac{1}{u^{1/2}}du$$ $$A=\frac{1}{3}\int u^{-1/2}du$$ $$A=\frac{1}{3}\frac{u^{1/2}}{\frac{1}{2}}+C$$ $$A=\frac{\sqrt u}{\frac{3}{2}}+C$$ $$A=\frac{2\sqrt u}{3}+C$$ $$A=\frac{2\sqrt{3ax+bx^3}}{3}+C$$
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