Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 67


$$\frac{1}{2} \sec^{-1}x^2+c.$$

Work Step by Step

Let $u=x^2$, then $du=2xdx$ and hence we have $$\int \frac{xdx}{x\sqrt{x^4-1}}=\frac{1}{2}\int \frac{du}{u\sqrt{u^2-1}}=\frac{1}{2} \sec^{-1}u+c\\ =\frac{1}{2} \sec^{-1}x^2+c.$$
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