1. **State the problem:** We need to find the integral $$\int \frac{1}{x^3 (x^2 + 1)^2} \, dx.$$\n\n2. **Formula and approach:** To integrate rational functions like this, we use partial fraction decomposition. The denominator factors as $x^3$ and $(x^2+1)^2$.\n\n3. **Set up partial fractions:** We write\n$$\frac{1}{x^3 (x^2 + 1)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx + E}{x^2 + 1} + \frac{Fx + G}{(x^2 + 1)^2}.$$\n\n4. **Multiply both sides by the denominator:**\n$$1 = A x^2 (x^2 + 1)^2 + B x (x^2 + 1)^2 + C (x^2 + 1)^2 + (Dx + E) x^3 (x^2 + 1) + (Fx + G) x^3.$$\n\n5. **Expand and collect terms:** This is lengthy, but equate coefficients of powers of $x$ to solve for constants $A,B,C,D,E,F,G$.\n\n6. **Shortcut by plugging values:** Plug $x=0$ to find $C$:\n$$1 = C (0^2 + 1)^2 = C \implies C=1.$$\n\n7. **After solving system (omitted detailed algebra here), the partial fractions are:**\n$$\frac{1}{x^3 (x^2 + 1)^2} = \frac{1}{x^3} - \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{1}{(x^2 + 1)^2}.$$\n\n8. **Integrate each term separately:**\n$$\int \frac{1}{x^3} dx = \int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2 x^2} + C_1,$$\n$$\int -\frac{1}{x} dx = -\ln|x| + C_2,$$\n$$\int \frac{x}{x^2 + 1} dx.$$ Use substitution $u = x^2 + 1$, $du = 2x dx$, so\n$$\int \frac{x}{x^2 + 1} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln(x^2 + 1) + C_3,$$\n$$\int -\frac{1}{(x^2 + 1)^2} dx.$$ This integral is known: $$\int \frac{dx}{(x^2 + 1)^2} = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan x + C_4,$$ so\n$$\int -\frac{1}{(x^2 + 1)^2} dx = -\frac{x}{2(x^2 + 1)} - \frac{1}{2} \arctan x + C_5.$$\n\n9. **Combine all results:**\n$$\int \frac{1}{x^3 (x^2 + 1)^2} dx = -\frac{1}{2 x^2} - \ln|x| + \frac{1}{2} \ln(x^2 + 1) - \frac{x}{2(x^2 + 1)} - \frac{1}{2} \arctan x + C.$$