Answer
$(\frac{r^3}{18}-1)^6+C$
Work Step by Step
Let u=$\frac{r^3}{18}-1=>du=\frac{r^2}{6}dr=>6du=r^2dr$
=$\int (\frac{r^3}{18}-1)^5dr$
=$\int u^5(6du)=6 \int u^5du$
=$6(\frac{u^6}{6})+C$
=$(\frac{r^3}{18}-1)^6+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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: 27
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