1. **State the problem:** Decompose the rational function $$\frac{x^2+x+1}{(x-1)(x+2)}$$ into partial fractions. 2. **Formula and rules:** For distinct linear factors in the denominator, the partial fraction decomposition takes the form: $$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$ where $A$ and $B$ are constants to be determined. 3. **Set up the equation:** $$\frac{x^2+x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ Multiply both sides by $(x-1)(x+2)$ to clear denominators: $$x^2 + x + 1 = A(x+2) + B(x-1)$$ 4. **Expand the right side:** $$x^2 + x + 1 = A x + 2A + B x - B$$ Combine like terms: $$x^2 + x + 1 = (A + B) x + (2A - B)$$ 5. **Equate coefficients:** - Coefficient of $x^2$: Left side has 1, right side has 0, so $1 = 0$ which is impossible. This means the numerator degree is not less than denominator degree, so first perform polynomial division. 6. **Polynomial division:** Divide $x^2 + x + 1$ by $(x-1)(x+2) = x^2 + x - 2$. Since degrees are equal, divide: $$\frac{x^2 + x + 1}{x^2 + x - 2} = 1 + \frac{3}{x^2 + x - 2}$$ 7. **Now decompose the remainder:** $$\frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ Multiply both sides by $(x-1)(x+2)$: $$3 = A(x+2) + B(x-1)$$ 8. **Expand and group:** $$3 = A x + 2A + B x - B = (A + B) x + (2A - B)$$ 9. **Equate coefficients:** - Coefficient of $x$: $0 = A + B$ - Constant term: $3 = 2A - B$ 10. **Solve the system:** From $0 = A + B$, we get $B = -A$. Substitute into $3 = 2A - B$: $$3 = 2A - (-A) = 3A \implies A = 1$$ Then $B = -1$. 11. **Write the final decomposition:** $$\frac{x^2 + x + 1}{(x-1)(x+2)} = 1 + \frac{1}{x-1} - \frac{1}{x+2}$$