How do I find the partial fraction decomposition of a rational function?

Here is the systematic approach using the Heaviside cover-up method for non-repeated factors and substitution for repeated factors. **Step 1: Set up the decomposition.** $$\frac{x^2 + 2x + 3}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$ **Step 2: Find $A$ using the cover-up method.** Multiply both sides by $(x-1)$ and evaluate at $x = 1$: $$A = \left.\frac{x^2 + 2x + 3}{(x+2)^2}\right|_{x=1} = \frac{1 + 2 + 3}{(3)^2} = \frac{6}{9} = \frac{2}{3}$$ **Step 3: Find $C$ using the cover-up method for the repeated factor.** Multiply both sides by $(x+2)^2$ and evaluate at $x = -2$: $$C = \left.\frac{x^2 + 2x + 3}{x-1}\right|_{x=-2} = \frac{4 - 4 + 3}{-3} = \frac{3}{-3} = -1$$ **Step 4: Find $B$ by substituting a convenient value.** We have $A = 2/3$ and $C = -1$. Substitute $x = 0$ into the original equation: $$\frac{3}{(-1)(2)^2} = \frac{2/3}{-1} + \frac{B}{2} + \frac{-1}{4}$$ $$-\frac{3}{4} = -\frac{2}{3} + \frac{B}{2} - \frac{1}{4}$$ $$\frac{B}{2} = -\frac{3}{4} + \frac{2}{3} + \frac{1}{4} = -\frac{1}{2} + \frac{2}{3} = \frac{1}{6} \quad \Rightarrow \quad B = \frac{1}{3}$$ **Step 5: Write the decomposition.** $$\frac{x^2 + 2x + 3}{(x-1)(x+2)^2} = \frac{2/3}{x-1} + \frac{1/3}{x+2} - \frac{1}{(x+2)^2}$$ **General Strategy:** | Factor Type | Term Form | |---|---| | Distinct linear $x - a$ | $\frac{A}{x - a}$ | | Repeated linear $(x - a)^k$ | $\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k}$ | | Irreducible quadratic $x^2 + bx + c$ | $\frac{Ax + B}{x^2 + bx + c}$ | | Repeated quadratic $(x^2+bx+c)^k$ | $\frac{A_1x+B_1}{x^2+bx+c} + \cdots + \frac{A_kx+B_k}{(x^2+bx+c)^k}$ |