Prove

Let $1+2 x^{2}=t$

$\therefore 4 x d x=d t$

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

$=\frac{1}{4} \int t^{\frac{1}{2}} d t$

$=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$

$=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$