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: 27

Answer

$$\int(x^2+1)(x^3+3x)^4dx=\frac{(x^3+3x)^5}{15}+C$$

Work Step by Step

$$A=\int(x^2+1)(x^3+3x)^4dx$$ Let $u=x^3+3x$. We would have $du=3x^2+3dx=3(x^2+1)dx$. Therefore, $(x^2+1)dx=\frac{1}{3}du$ Substitute into $A$, we have $$A=\frac{1}{3}\int u^4du$$ $$A=\frac{1}{3}\frac{u^5}{5}+C$$ $$A=\frac{u^5}{15}+C$$ $$A=\frac{(x^3+3x)^5}{15}+C$$
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