Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 31

Answer

$$\frac{{{6^{{x^3} + 8}}}}{{3\ln 6}} + C$$

Work Step by Step

$$\eqalign{ & \int {{x^2}{6^{{x^3} + 8}}dx} \cr & {\text{substitute }}u = {x^3} + 8,{\text{ }}du = 3{x^2}dx \cr & \int {{x^2}{6^{{x^3} + 8}}dx} = \frac{1}{3}\int {{6^u}du} \cr & {\text{find the antiderivative}} \cr & {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & {\text{letting }}a = 6 \cr & = \frac{{{6^u}}}{{3\ln 6}} + C \cr & = \frac{{{6^{{x^3} + 8}}}}{{3\ln 6}} + C \cr} $$
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