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: 24


$$\int x\sqrt {x+2}dx=\frac{2}{5}\sqrt{(x+2)^5}-\frac{4}{3}\sqrt{(x+2)^3}+C$$

Work Step by Step

$$A=\int x\sqrt {x+2}dx$$ Let $u=\sqrt{x+2}$. We would have $u^2=x+2$. Therefore, $(u^2)'du=(x+2)'dx$. That means $2udu=dx$ Also, since $u^2=x+2$, so $x=u^2-2$ Substitute into $A$, we have $$A=\int(u^2-2)u\times2udu$$ $$A=2\int(u^2-2)u^2du$$ $$A=2\int(u^4-2u^2)du$$ $$A=2(\frac{u^5}{5}-\frac{2u^3}{3})+C$$ $$A=\frac{2u^5}{5}-\frac{4u^3}{3}+C$$ $$A=\frac{2}{5}\sqrt{(x+2)^5}-\frac{4}{3}\sqrt{(x+2)^3}+C$$
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