Integrate the function $x \sqrt{1+2 x^{2}}$.

(N/A) Let $1+2 x^{2} = t$.
Then,differentiating both sides with respect to $x$,we get $4x \, dx = dt$,which implies $x \, dx = \frac{dt}{4}$.
Substituting these into the integral:
$\int x \sqrt{1+2 x^{2}} \, dx = \int \sqrt{t} \cdot \frac{dt}{4} = \frac{1}{4} \int t^{1/2} \, dt$.
Using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$:
$= \frac{1}{4} \left( \frac{t^{3/2}}{3/2} \right) + C = \frac{1}{4} \cdot \frac{2}{3} t^{3/2} + C = \frac{1}{6} t^{3/2} + C$.
Substituting back $t = 1+2x^2$:
$= \frac{1}{6} (1+2x^2)^{3/2} + C$,where $C$ is an arbitrary constant.