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