University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 42



Work Step by Step

Given: $\int \dfrac{4 +\sqrt t}{t^3} dt$ Since, we have $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, $\int \dfrac{4 +\sqrt t}{t^3} dt=\int \dfrac{4}{t^{3}}+\dfrac{t^{1/2}}{t^3} dt=4(\dfrac{1}{-2}t^{-2}+\dfrac{1}{-3/2}t^{-3/2}+C=-\dfrac{2}{t^2}-\dfrac{2}{3t^{3/2}}+C$
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