Answer
$$\frac{1}{2}\sin^{-1}t^2+c.$$
Work Step by Step
Let $u=t^2$, then $du=2t \ dt$ and hence we get
$$\int \frac{t}{\sqrt{1-t^4}}dt= \frac{1}{2}\int \frac{1}{\sqrt{1-u^2}}du \\
=\frac{1}{2}\sin^{-1}u+c=\frac{1}{2}\sin^{-1}t^2+c.$$
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: 108
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