1. We are asked to find the primitive (antiderivative) of the function $$f(x) = \frac{x^3 + 2x}{\sqrt{x^4 + 4x^2 + 2}}$$ over the real numbers. 2. To find the primitive, we look for an expression $$F(x)$$ such that $$F'(x) = f(x)$$. 3. Notice the denominator is $$\sqrt{x^4 + 4x^2 + 2}$$. Let's set $$u = x^4 + 4x^2 + 2$$. 4. Then, $$\frac{du}{dx} = 4x^3 + 8x$$. 5. The numerator is $$x^3 + 2x$$, which is exactly half of $$4x^3 + 8x$$, so $$x^3 + 2x = \frac{1}{4} \frac{du}{dx}$$. 6. Rewrite the integral of $$f(x)$$ as: $$\int \frac{x^3 + 2x}{\sqrt{u}} dx = \int \frac{1}{4} \frac{du/dx}{\sqrt{u}} dx = \frac{1}{4} \int \frac{du}{\sqrt{u}}$$ 7. The integral $$\int u^{-1/2} du = 2 \sqrt{u} + C$$. 8. Therefore, $$\int f(x) dx = \frac{1}{4} \times 2 \sqrt{u} + C = \frac{1}{2} \sqrt{x^4 + 4x^2 + 2} + C$$ 9. Final answer: $$F(x) = \frac{1}{2} \sqrt{x^4 + 4x^2 + 2} + C$$