1. **State the problem:** Rewrite the rational function $$\frac{x^2 + x + 4}{x (x - 2)^2}$$ as partial fractions. 2. **Set up the partial fraction decomposition:** Since the denominator factors are $x$ and $(x-2)^2$, the decomposition form is: $$\frac{x^2 + x + 4}{x (x - 2)^2} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$$ 3. **Multiply both sides by the denominator to clear fractions:** $$x^2 + x + 4 = A (x - 2)^2 + B x (x - 2) + C x$$ 4. **Expand the right side:** $$A (x^2 - 4x + 4) + B x (x - 2) + C x = A x^2 - 4 A x + 4 A + B x^2 - 2 B x + C x$$ 5. **Group like terms:** $$x^2 + x + 4 = (A + B) x^2 + (-4 A - 2 B + C) x + 4 A$$ 6. **Equate coefficients of corresponding powers of $x$:** - Coefficient of $x^2$: $1 = A + B$ - Coefficient of $x$: $1 = -4 A - 2 B + C$ - Constant term: $4 = 4 A$ 7. **Solve for $A$ from the constant term:** $$4 = 4 A \implies A = 1$$ 8. **Substitute $A=1$ into the $x^2$ coefficient equation:** $$1 = 1 + B \implies B = 0$$ 9. **Substitute $A=1$ and $B=0$ into the $x$ coefficient equation:** $$1 = -4(1) - 2(0) + C \implies 1 = -4 + C \implies C = 5$$ 10. **Write the final partial fraction decomposition:** $$\frac{x^2 + x + 4}{x (x - 2)^2} = \frac{1}{x} + \frac{0}{x - 2} + \frac{5}{(x - 2)^2} = \frac{1}{x} + \frac{5}{(x - 2)^2}$$ **Answer:** $$\boxed{\frac{1}{x} + \frac{5}{(x - 2)^2}}$$