Integrate $\int{\frac{x}{1+x^4}}dx$

$$\int \frac{x}{1+x^4}\, dx=\frac{1}{2}\int \frac{1}{1+\left(x^2\right)^2}\, d\left(x^2\right)=\frac{1}{2}\arctan x^2+C$$

$$\int{\frac{1}{1+x^2}dx}=\mathrm{arctan}(x)$$ The above is a standard integral. Using that, with $y=x^2$, thus $dy=2xdx$, we find: $$\int{\frac{xdx}{1+x^4}}=\frac{1}{2}\int{\frac{2xdx}{1+\left(x^2\right)^2}}=\frac{1}{2}\int{\frac{dy}{1+y^2}}=\frac{1}{2}\mathrm{arctan}(y)=\frac{1}{2}\mathrm{arctan}(x^2)$$