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.