Prove

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$\Rightarrow \int \frac{3 x^{2}}{x^{6}+1} d x=\int \frac{d t}{t^{2}+1}$

$=\tan ^{1} t+C$

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