Evaluate the integral: I = integral (x³)/(sqrt(1+2x²)) dx

Evaluate the integral: $$ I = \int \frac{x^3}{\sqrt{1+2x^2}}\,dx. $$

Explanation of Method

The substitution $u=1+2x^2$ turns $x^3\,dx=x^2\cdot x\,dx$ into $\tfrac14(u-1)\,du$ over $\sqrt u$, a direct power-rule integral.

Let $u=1+2x^2$, so $du=4x\,dx$ and $x^2=(u-1)/2$:

$$ I = \int\frac{x^2\cdot x\,dx}{\sqrt u} = \frac18\int\frac{u-1}{\sqrt u}\,du = \frac18\int\left(u^{1/2}-u^{-1/2}\right)du. $$

$$ I = \frac18\left[\frac{2u^{3/2}}3-2u^{1/2}\right] + C = \frac{u^{3/2}}{12} - \frac{u^{1/2}}4 + C. $$

With $u=1+2x^2$, factoring out $\sqrt u/12$:

Piece	Antiderivative
$u^{1/2}/8$	$u^{3/2}/12$
$-u^{-1/2}/8$	$-u^{1/2}/4$

The domain is all real $x$, since $1+2x^2>0$ everywhere.
The result was verified by differentiation at $x=1$: the derivative matches the integrand $x^3/\sqrt{1+2x^2}$ exactly.

The solution to the integral is:

$$ I = \frac{\left(x^2-1\right)\sqrt{1+2x^2}}6 + C. $$