Answer
\[\frac{1}{2}\sin^{-1} (x^2)+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{x}{\sqrt {1-x^4}}\]
\[\Rightarrow I=\frac{1}{2}\int\frac{2x}{\sqrt {1-x^4}}\]
Put \[t=x^2\;\;\;...(1)\]
\[\Rightarrow dt=2x \,dx\]
\[I=\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}}\]
\[I=\frac{1}{2}\sin^{-1} (t)+C\]
Where $C$ is constant of integration
Using (1)
\[I=\frac{1}{2}\sin^{-1} (x^2)+C\]
Hence , \[\int\frac{x}{\sqrt {1-x^4}}=\frac{1}{2}\sin^{-1} (x^2)+C\]
