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: 3


$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$

Work Step by Step

$$A=\int x^2\sqrt{x^3+1}dx$$ Let $u=x^3+1$ Then $du=(x^3+1)'dx=3x^2dx$. So $x^2dx=\frac{1}{3}du$. Substitute into $A$, we have $$A=\int \sqrt u(\frac{1}{3})du$$ $$A=\frac{1}{3}\int u^{1/2}du$$ $$A=\frac{1}{3}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{u^{3/2}}{\frac{9}{2}}+C$$ $$A=\frac{2\sqrt{u^3}}{9}+C$$ $$A=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$
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