Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 94

Answer

$=\frac{5}{4}f^4(x)+\frac{7}{3}f^3(x)+\frac{1}{2}f^2(x)+c$

Work Step by Step

$\int(5f^3(x)+7f^2(x)+f(x))f'(x)dx$ $u=f(x)$ $dx=\frac{du}{f'(x)}$ $\int(5u^3+7u^2+u)du=\int5u^3du+\int7u^2du+\int u*du$ $=\frac{5}{4}u^4+\frac{7}{3}u^3+\frac{1}{2}u^2+c$ $=\frac{5}{4}f^4(x)+\frac{7}{3}f^3(x)+\frac{1}{2}f^2(x)+c$
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