Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 34

Answer

$$ \int \frac{x}{\sqrt{1-x^{4}}} d x= \frac{1}{2} \sin ^{-1}\left(x^{2}\right)+C $$ where $ C $ is an arbitrary constant.

Work Step by Step

$$ \int \frac{x}{\sqrt{1-x^{4}}} d x $$ Let $ u=x^{2}$. Then $ d u=2 x d x, $ so $$ \begin{split} \int \frac{x}{\sqrt{1-x^{4}}} d x & =\frac{1}{2} \int \frac{d u}{\sqrt{1-u^{2}}} \\ &=\frac{1}{2} \sin ^{-1} u+C \\ &=\frac{1}{2} \sin ^{-1}\left(x^{2}\right)+C \end{split} $$ where $ C $ is an arbitrary constant.
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