University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 99



Work Step by Step

Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Now, $\int (3 \sqrt t+\dfrac{4}{t^2}) dt=\int (3 t^{1/2}+4t^{-2}) dt=\dfrac{3t^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1}+ \dfrac{4t^{-2+1}}{-2+1}+c$ or, $=2t^{3/2}-\dfrac{4}{t}+c$
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