1. **State the problem:** Evaluate the integral $$\int \frac{dx}{x^2 (x^2 + 1)}.$$\n\n2. **Formula and approach:** We use partial fraction decomposition for rational functions. The integrand can be decomposed as $$\frac{1}{x^2 (x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}.$$\n\n3. **Set up the equation:** Multiply both sides by the denominator $$x^2 (x^2 + 1)$$ to clear fractions:\n$$1 = A x (x^2 + 1) + B (x^2 + 1) + (Cx + D) x^2.$$\n\n4. **Expand and collect terms:**\n$$1 = A x^3 + A x + B x^2 + B + C x^3 + D x^2.$$\nGroup by powers of $$x$$:\n$$1 = (A + C) x^3 + (B + D) x^2 + A x + B.$$\n\n5. **Equate coefficients:** Since the left side is constant 1, coefficients of $$x^3, x^2, x$$ must be zero, and constant term must be 1:\n$$\begin{cases} A + C = 0 \\ B + D = 0 \\ A = 0 \\ B = 1 \end{cases}$$\n\n6. **Solve the system:** From $$A=0$$, then $$C = -A = 0$$. From $$B=1$$, then $$D = -B = -1$$.\n\n7. **Rewrite the decomposition:**\n$$\frac{1}{x^2 (x^2 + 1)} = \frac{0}{x} + \frac{1}{x^2} + \frac{0 \cdot x - 1}{x^2 + 1} = \frac{1}{x^2} - \frac{1}{x^2 + 1}.$$\n\n8. **Integrate term-by-term:**\n$$\int \frac{dx}{x^2} - \int \frac{dx}{x^2 + 1} = \int x^{-2} dx - \int \frac{dx}{x^2 + 1}.$$\n\n9. **Compute each integral:**\n$$\int x^{-2} dx = \int x^{-2} dx = -x^{-1} + C_1 = -\frac{1}{x} + C_1,$$\n$$\int \frac{dx}{x^2 + 1} = \arctan x + C_2.$$\n\n10. **Combine results:**\n$$\int \frac{dx}{x^2 (x^2 + 1)} = -\frac{1}{x} - \arctan x + C,$$ where $$C = C_1 - C_2$$ is the constant of integration.\n\n**Final answer:** $$\boxed{\int \frac{dx}{x^2 (x^2 + 1)} = -\frac{1}{x} - \arctan x + C}.$$