Prove

Let $\sqrt{2} x^{2}=t$

$\therefore 2 \sqrt{2} x d x=d t$

$\Rightarrow \int \frac{3 x}{1+2 x^{4}} d x=\frac{3}{2 \sqrt{2}} \int \frac{d t}{1+t^{2}}$

$=\frac{3}{2 \sqrt{2}}\left[\tan ^{-1} t\right]+\mathrm{C}$

$=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2} x^{2}\right)+\mathrm{C}$