Answer
$$\frac{1}{2} \sec^{-1}x^2+c.$$
Work Step by Step
Let $u=x^2$, then $du=2xdx$ and hence we have
$$\int \frac{xdx}{x\sqrt{x^4-1}}=\frac{1}{2}\int \frac{du}{u\sqrt{u^2-1}}=\frac{1}{2} \sec^{-1}u+c\\
=\frac{1}{2} \sec^{-1}x^2+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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