Answer
$$\int\frac{x}{1+x^4}dx=\frac{1}{2}\arctan(x^2)+c$$
Work Step by Step
To evaluate the integral
$$\int\frac{x}{1+x^4}dx$$
we will use substitution $x^2=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$ . Putting this into the integral we get:
$$\int\frac{x}{1+x^4}dx=\int\frac{1}{1+t^2}\frac{dt}{2}=\frac{1}{2}\arctan t+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int\frac{x}{1+x^4}dx=\frac{1}{2}\arctan t+c=\frac{1}{2}\arctan(x^2)+c$$