Answer
$$
\int(3t-4 )^{5} dt
=\frac{1}{18} (3t-4)^6+c.
$$
Work Step by Step
Since $ u=3t-4 $, then $ du=3dt $ and hence, $$
\int(3t-4 )^{5} dt=\frac{1}{3}\int u^{5} d u=\frac{1}{3} \frac{1}{6}u^6+c\\
=\frac{1}{18} (3t-4)^6+c.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 9
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