University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 279: 119


$\dfrac{3}{2} \sec^{-1} x+C$

Work Step by Step

Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and where $c$ is a constant of proportionality. Then $\int \dfrac{3}{2x\sqrt {x^2-1}} dx=(\dfrac{3}{2})\int \dfrac{1}{x\sqrt {x^2-1}} dx$ As we know that $ \int \dfrac{3}{2x\sqrt {x^2-1}} dx=\sec^{-1} x+C$ Thus, $= \dfrac{3}{2} \sec^{-1} x+C$
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