Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Basic Idea

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

MathJax TeX Test Page For example, you can turn $\dfrac{8x + 8}{x^2 + 4x - 12}$ into $\dfrac{3}{x-2} + \dfrac{5}{x+6}$.

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

MathJax TeX Test Page For example, you can turn $\dfrac{8x + 8}{x^2 + 4x - 12}$ into $\dfrac{3}{x-2} + \dfrac{5}{x+6}$.

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

MathJax TeX Test Page $$\dfrac{19-3x}{(x-4)^2}$$ If we factor the denominator we only have $(x-4)$, so what we have to do is to have a fraction with just a number divided by $(x-4)^2$. It will look like this: $$\dfrac{19-3x}{(x-4)^2} = \frac{A}{x-4} + \frac{B}{(x-4)^2}$$ We do the same thing as before: multiply through by $(x-4)^2$ $$19-3x = A(x-4) + B$$ When $x = 4, B = 7$. From that, you can see that $12 - 3x = A(x-4)$, so A = -3. However, there may be other roots, so in general, you plug in a constant. Let's plug in $x = 0$, so $19 = -4A + 7 \to A = -3$. $$\dfrac{19-3x}{(x-4)^2} = \boxed{\frac{7}{(x-4)^2}-\frac{3}{x-4}}$$

Other Example

MathJax TeX Test Page $$\int\frac{x^2 - 2x - 37}{x^2 - 3x - 40}\,\mathrm{dx}$$ Before we do this, we must remember our only rule: the degree of the numerator is less than the denominator. So, we must turn that into a regular fraction. $$\frac{x^2 - 2x - 37}{x^2 - 3x - 40} = \frac{x^2 - 3x - 40 + x + 3}{x^2 - 3x - 40} = 1 + \frac{x+3}{x^2 - 3x - 40}$$ Now we can proceed. $$\frac{x+3}{x^2 - 3x - 40} = \frac{A}{x+5} + \frac{B}{x-8}$$ $$x+3 = A(x-8) + B(x+5)$$ $$x = 8 \to B = \frac{11}{13}, x = -5 \to A = \frac{2}{13}$$ If you notice, we got the same results. The reason for this is we plugged in $x = -5$ and $x = 8$, both roots. So, $x^2 - 2x - 37 = x^2 - 3x - 40 + x + 3$, and plugging in $-5 or 8$ would make the first part 0 leaving $x + 3$. If you plug in anything else, it'll be different. $$\int\frac{x^2 - 2x - 37}{x^2 - 3x - 40}\,\mathrm{dx} = \int 1 + \dfrac{\frac{2}{13}}{x+5} + \dfrac{\frac{11}{13}}{x - 8}\,\mathrm{dx}$$$$= \boxed{x + \frac{2}{13}\ln|x + 5| + \frac{11}{13}\ln|x-8| + C}$$ The moral of the story is to ensure that the fraction is a proper fraction.
David Witten