University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 21


$\sin^{-1} x-\sqrt {1-x^2}+C $

Work Step by Step

Since, we have $\int \sqrt{\dfrac{x+1}{1-x}} dx$ Now, the given integral can be written as: $\int \sqrt{\dfrac{x+1}{1-x}} dx =\int \dfrac{x}{\sqrt{1-x^2}} dx +\int \dfrac{1}{\sqrt{1-x^2}} dx $ $=-\sqrt {1-x^2}+\sin^{-1} x+C$ or, $=\sin^{-1} x-\sqrt {1-x^2}+C $
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