So you've never been partial to fractions.

Partial Fraction Decomposition (PFD here for short) has been known to come as a surprising blast from the past for students in Calculus II and Differential Equations as something they (should) have learned in Algebra. Whether the student simply forgot they were taught it (most often the case) or the instructor failed to cover it (the most blamed case :ahem:) it needs to be understood regardless and is quite a nifty technique to tackle a variety of problems.
The basic idea of PFD is to reverse the process of adding two or more fractions, recovering the summands.
The first step is to completely factor the denominator of the fraction you wish to decompose. The example I am going to use will have all three of the cases that can crop up in a PFD you're likely to see!
Behold:
\[ \frac{2x+9}{x(x+1)(x^2+2)(x2)^3} \]
Now, obviously I've already factored the bottom and you're not likely to see one this ugly on a regular basis (but it's a good example).
Let's check out the three cases of denominator factors and what we're going to do with them.
The basic idea of PFD is to reverse the process of adding two or more fractions, recovering the summands.
The first step is to completely factor the denominator of the fraction you wish to decompose. The example I am going to use will have all three of the cases that can crop up in a PFD you're likely to see!
Behold:
\[ \frac{2x+9}{x(x+1)(x^2+2)(x2)^3} \]
Now, obviously I've already factored the bottom and you're not likely to see one this ugly on a regular basis (but it's a good example).
Let's check out the three cases of denominator factors and what we're going to do with them.
DISTINCT LINEAR FACTORS
A distinct linear factor is simply a linear (1st power) factor that only appears once. Notice in our example that that includes the factors \( x \) and \( (x+1) \)
Our response to one of these is to write a sum including a fraction for each factor with a variable constant coefficient in the numerator: \[ \frac{A}{x} + \frac{B}{x+1} \] 
REPEATED LINEAR FACTORS
A repeated linear factor is a factor that appears to a higher power than one. Notice in the example that this includes only the factor \( (x2)^3 \).
In these cases, what we do is write a sum of fractions, one for each power of the factor in question, again with a variable constant in the numerator: \[ \frac{A}{x2} + \frac{B}{(x2)^2} + \frac{C}{(x2)^3} \] 
PRIME QUADRATIC FACTORS
A prime quadratic factor is one that cannot be broken down any further in the Real Numbers. In this example, the only prime quadratic factor is \( (x^2+2) \).
In this last case, our response is to write a sum including a fraction for each prime quadratic factor with a linear expression in the numerator: \[ \frac{Ax+B}{x^2+2} \] 
These rules taken together (without repeating variable constants twice, don't do that!) gives us the following response to the example:
\[ \frac{2x+9}{x(x+1)(x^2+2)(x2)^3} = \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x2}+\frac{D}{(x2)^2}+\frac{E}{(x2)^3}+\frac{Fx+G}{x^2+2} \]
\[ \frac{2x+9}{x(x+1)(x^2+2)(x2)^3} = \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x2}+\frac{D}{(x2)^2}+\frac{E}{(x2)^3}+\frac{Fx+G}{x^2+2} \]
Ok, so now that we know how to set up the decomposition, let's check out a few easier to manage problems.
\[ \frac{x+1}{x^3+x} \]
First, we need to factor the denominator to check out how many of each case we have.
\[ \frac{x+1}{x(x^2+1)} \]
This contains a distinct linear factor ( \( x \) ) and a prime quadratic factor ( \( x^2+1 \) ), so the set up would be the following:
\[ \frac{x+1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \]
Now that we're all set up, what happens next?
Multiply both sides of the equation by the denominator of the left side (the original fraction).
This kills all the fractions and paves the road for figuring out what numbers A, B, and C need to be.
\[ x+1 = A(x^2+1)+(Bx+C)x \]
Or, simplifying and gathering like terms...
\[ x+1 = (A+B)x^2+Cx+A \]
We can now compare the left and right hand side to figure out our variables.
\[ \frac{x+1}{x^3+x} \]
First, we need to factor the denominator to check out how many of each case we have.
\[ \frac{x+1}{x(x^2+1)} \]
This contains a distinct linear factor ( \( x \) ) and a prime quadratic factor ( \( x^2+1 \) ), so the set up would be the following:
\[ \frac{x+1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \]
Now that we're all set up, what happens next?
Multiply both sides of the equation by the denominator of the left side (the original fraction).
This kills all the fractions and paves the road for figuring out what numbers A, B, and C need to be.
\[ x+1 = A(x^2+1)+(Bx+C)x \]
Or, simplifying and gathering like terms...
\[ x+1 = (A+B)x^2+Cx+A \]
We can now compare the left and right hand side to figure out our variables.
This is called comparing coefficients below, and relies on the fact that when two equations are equal, they have to have the exact same amount of each 'unlike' term, because I can't make a higher power of \( x \) out of numerical coefficients!
There are no \( x^2 \) on the left:
That means there can't be any on the right, and hence \( A+B = 0 \). 
There is a single \( x \) on the left:
That means there is only one on the right, and so \( C=1 \). 
The constant term on the left is \( 1 \):
The constant term on the right has to be the same, so \( A = 1 \), making \( B = 1 \). 
Now that we know what A,B, and C are, we just have to plug them in our "set up" phase and we'll be done.
\[ \frac{x+1}{x(x^2+1)} = \frac{1}{x} + \frac{x+1}{x^2+1} \]
You can add them back together and verify if you'd like. These are easy to double check.
\[ \frac{x+1}{x(x^2+1)} = \frac{1}{x} + \frac{x+1}{x^2+1} \]
You can add them back together and verify if you'd like. These are easy to double check.
Here's one more quick example:
\[ \frac{3x^2+x1}{x^3x^2} \]
1. Factor the denominator:
\[ \frac{3x^2+x1}{x^2(x1)} \]
2. Set up according to the cases:
\[ \frac{3x^2+x1}{x^2(x1)} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x1} \]
3. Multiply through by the denominator of the original:
\[ 3x^2+x1 = Ax(x1)+B(x1)+Cx^2 \]
4. Simplify and gather like terms:
\[ 3x^2+x1 = (A+C)x^2+(BA)xB \]
5. Compare coefficients:
We see that B=1, A=0, C=3
6. Plug back in to the 'set up':
\[ \frac{1}{x^2}+\frac{3}{x1} \]
Done.
\[ \frac{3x^2+x1}{x^3x^2} \]
1. Factor the denominator:
\[ \frac{3x^2+x1}{x^2(x1)} \]
2. Set up according to the cases:
\[ \frac{3x^2+x1}{x^2(x1)} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x1} \]
3. Multiply through by the denominator of the original:
\[ 3x^2+x1 = Ax(x1)+B(x1)+Cx^2 \]
4. Simplify and gather like terms:
\[ 3x^2+x1 = (A+C)x^2+(BA)xB \]
5. Compare coefficients:
We see that B=1, A=0, C=3
6. Plug back in to the 'set up':
\[ \frac{1}{x^2}+\frac{3}{x1} \]
Done.