Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 81


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

Work Step by Step

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