# Basic Idea

The purpose of this method is to make one complicated polynomial fraction into multiple simple ones.

Why is this better? Mainly, it's much easier to work with. $$\int\dfrac{8x + 8}{x^2 + 4x - 12}\, \mathrm{d}x \text{ is much harder than } \int \dfrac{3}{x-2} + \dfrac{5}{x+6}\, \mathrm{d}x$$ On the right, we can instantly see that it's $\boxed{3\ln|x-2| + 5\ln|x + 6| + C}$.

# Method

Let's use the last example. We begin with this: $$\dfrac{8x + 8}{x^2 + 4x - 12}$$ If we factor the denominator we get $(x-2)(x+6)$. The first important rule is that the degree of the polynomial in the numerator is

**less than**the degree of the denominator. $$\dfrac{8x + 8}{x^2 + 4x - 12} = \dfrac{A}{x-2} + \frac{B}{x+6}$$ If we multiply through by $x^2 + 4x - 12$, we get $$8x + 8 = A(x + 6) + B(x-2)$$ Now, we can solve for A and B. We can do this in two ways. First, we can combine the terms and get $$8x + 8 = (A + B)x + 6A - 2B$$ Because x and 1 are linearly independent, we can solve a system of equations for A and B: $$\begin{cases}8 = A + B\\8 = 6A - 2B\end{cases}$$ There's a better way! If we start with $8x + 8 = A(x + 6) + B(x-2)$, we can plug in $x = -6$ and $x = 2$ and solve for A and B that way. $$x = -6 \to -40 = -8B \to B = 5$$ $$x = 2 \to 24 = 8A \to A = 3$$ $$\dfrac{8x + 8}{x^2 + 4x - 12} = \dfrac{A}{x-2} + \frac{B}{x+6} = \boxed{\dfrac{3}{x-2} + \frac{5}{x+6}}$$

# Repeated Root Example

# Other Example

David Witten