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Partial Fractions

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

Work

Integrating Polar and Parametric Functions