1. **State the problem:** We need to evaluate the integral $$\int \frac{x^3 + x}{x^4 + 2x^2} \, dx.$$\n\n2. **Simplify the integrand:** Factor the denominator and numerator where possible.\nThe denominator: $$x^4 + 2x^2 = x^2(x^2 + 2).$$\nThe numerator: $$x^3 + x = x(x^2 + 1).$$\nSo the integrand becomes $$\frac{x(x^2 + 1)}{x^2(x^2 + 2)} = \frac{x^2 + 1}{x(x^2 + 2)}.$$\n\n3. **Rewrite the integral:** $$\int \frac{x^2 + 1}{x(x^2 + 2)} \, dx.$$\n\n4. **Use partial fraction decomposition:** We want to express $$\frac{x^2 + 1}{x(x^2 + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2}.$$\nMultiply both sides by the denominator $$x(x^2 + 2):$$\n$$x^2 + 1 = A(x^2 + 2) + (Bx + C)x = A x^2 + 2A + B x^2 + C x.$$\nGroup like terms:\n$$x^2 + 1 = (A + B) x^2 + C x + 2A.$$\n\n5. **Equate coefficients:**\n- Coefficient of $$x^2$$: $$1 = A + B$$\n- Coefficient of $$x$$: $$0 = C$$\n- Constant term: $$1 = 2A$$\n\n6. **Solve the system:**\nFrom $$1 = 2A$$, we get $$A = \frac{1}{2}$$.\nFrom $$1 = A + B$$, substitute $$A$$: $$1 = \frac{1}{2} + B \Rightarrow B = \frac{1}{2}$$.\nFrom $$0 = C$$, we get $$C = 0$$.\n\n7. **Rewrite the integral with partial fractions:**\n$$\int \left( \frac{1/2}{x} + \frac{(1/2) x}{x^2 + 2} \right) dx = \frac{1}{2} \int \frac{1}{x} dx + \frac{1}{2} \int \frac{x}{x^2 + 2} dx.$$\n\n8. **Integrate each term:**\n- $$\int \frac{1}{x} dx = \ln|x| + C.$$\n- For $$\int \frac{x}{x^2 + 2} dx$$, use substitution $$u = x^2 + 2$$, so $$du = 2x dx$$ or $$\frac{du}{2} = x dx$$.\nThus, $$\int \frac{x}{x^2 + 2} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2 + 2| + C.$$\n\n9. **Combine results:**\n$$\frac{1}{2} \ln|x| + \frac{1}{2} \cdot \frac{1}{2} \ln|x^2 + 2| + C = \frac{1}{2} \ln|x| + \frac{1}{4} \ln|x^2 + 2| + C.$$\n\n**Final answer:** $$\int \frac{x^3 + x}{x^4 + 2x^2} \, dx = \frac{1}{2} \ln|x| + \frac{1}{4} \ln|x^2 + 2| + C.$$