Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 5


$\int (3x+2)(3x^2+4x)^4 dx=\frac{(3x^2+4x)^5}{10}+C$

Work Step by Step

$\int (3x+2)(3x^2+4x)^4 dx$ $\space$ and $\space \space u = 3x^2 + 4x$ $du = (6x +4)dx$ Doing the substitution $\space\space u = 3x^2 + 4x$ $\int (u)^4 \frac{du}{2}\space$ => $\space\frac{1}{2}\int u^4 du$ $\frac{1}{2}\int u^4 du = \frac{u^5}{10} + C$ Backing to x $\int (3x+2)(3x^2+4x)^4 dx=\frac{(3x^2+4x)^5}{10}+C$
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