Integrate the function: $\int \frac{3x}{1+2x^4} dx$

Let $\sqrt{2}x^2 = t$.
Then,differentiating both sides with respect to $x$,we get $2\sqrt{2}x dx = dt$,which implies $x dx = \frac{dt}{2\sqrt{2}}$.
Substituting these into the integral:
$\int \frac{3x}{1+2x^4} dx = 3 \int \frac{1}{1+t^2} \cdot \frac{dt}{2\sqrt{2}}$
$= \frac{3}{2\sqrt{2}} \int \frac{1}{1+t^2} dt$
$= \frac{3}{2\sqrt{2}} \tan^{-1}(t) + C$
$= \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C$,where $C$ is the constant of integration.