Answer
$$\frac{1}{2}\sin^{-1} x^2+c.$$
Work Step by Step
Let $u=x^2$, then $du=2xdx$. Now, we have
$$\int \frac{dx}{\sqrt{1-x^4}}=\frac{1}{2}\int \frac{du}{\sqrt{1-u^2}}\\
=\frac{1}{2}\sin u+c=\frac{1}{2}\sin^{-1} x^2+c.$$
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 - Chapter Review Exercises - Page 387: 81
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
