Partial Fraction Decomposition (PFD for short) is how we tackle complicated-looking rational functions. We can try this with any rational function where the degree of the numerator is smaller than the degree of the denominator (and all of the ones I'll show will fit that description). Let's start with an example: \[\int \dfrac{1}{x^3+2x^2+x} \ dx.\]

Thinking through the two methods we've seen so far, it doesn't look like this will yield to either u-substitution or integration by parts. Thus, we need a new method. However, as I keep pointing out, it really won't be a new method at all; what we're going to do is simply algebra. We're going to break this rational function down into several simpler fractions. Specifically, we'll find that \[\dfrac{1}{x^3+2x^2+x} = \dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2}.\]

Let's take a quick detour and prove that by getting a common denominator on those three fractions and combining them: \[\begin{align} \dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2} &= \dfrac{(x+1)^2}{x(x+1)^2} - \dfrac{x(x+1)}{x(x+1)^2} - \dfrac{x}{x(x+1)^2}\\ &= \dfrac{(x+1)^2-x(x+1)-x}{x(x+1)^2}\\ &= \dfrac{x^2+2x+1-x^2-x-x}{x(x+1)^2}\\ &= \dfrac{1}{x(x+1)^2} \end{align}\] Since, of course, \(x(x+1)^2 = x^3+2x^2+x\), we've proven that this combination of simpler fractions is equal to the more complicated one.

We don't know how to directly integrate \(\dfrac{1}{x^3+2x^2+x}\), but we can integrate \(\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2}\) by doing a quick u-substitution on each piece. Therefore, we've algebraically simplified the integral to the point where we can evaluate it: \[\begin{align} \int \dfrac{1}{x^3+2x^2+x} \ dx &= \int \dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2} \ dx\\ &= \ln x - \ln |x+1| + \dfrac{1}{x+1} + C \end{align}\] Notice that I've deliberately skipped the step of doing these simpler integrals with u-substitution; you should take a moment and verify this answer on your own.

This is all well and good, but the question, of course, is: how do we go about breaking something like \(\dfrac{1}{x^3+2x^2+x}\) down into something like \(\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2}\)? Well, let's look back at what we did when we combined the three fractions into one, and see if we can get some hint as to how to reverse the process. The first thing we notice is that the denominator was in factored form, and the denominators of the three simple fractions are similar to factors in the complicated denominator. Therefore, it shouldn't be a surprise that we'll start every partial fraction decomposition by factoring the denominator, and then we'll construct partial fractions with denominators similar to those factors.

In this example, we factor \(x^3+2x^2+x\) by first pulling out the \(x\) that's common to every term, and then breaking down the remaining quadratic: \[\begin{align}x^3+2x^2+x &= x(x^2+2x+1)\\ &= x(x+1)(x+1) \end{align}\]

This tells us what the *denominators* of the partial fractions will be, but not their numerators; we'll just have to label those as unknowns, and the rest of the problem is to find what they equal. Now, I'll give you clearer guidelines for this in a moment, but based on the factored form of that denominator, I know that the partial fractions will look as follows:
\[\dfrac{1}{x^3+2x^2+x} = \dfrac{1}{x(x+1)^2} = \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}\] The rest of the problem, as I said, is to find the values of \(A\), \(B\), and \(C\). There are two^{1} common ways to do this, but I'm going to just show you my preferred method, to avoid confusion.

Just like you would do to solve any rational equation, start by clearing the denominators. The way to do this is by multiplying both sides of the equation by the least common denominator, which is (and will be for every one of these problems) the full denominator on the left, since the denominators on the right are just individual chunks of that. Notice that when we do this, lots of things cancel on the right-hand side: \[\begin{align} \dfrac{1}{x(x+1)^2} &= \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}\\ x(x+1)^2 \cdot \dfrac{1}{x(x+1)^2} &= x(x+1)^2 \cdot \dfrac{A}{x} + x(x+1)^2 \cdot \dfrac{B}{x+1} + x(x+1)^2 \cdot \dfrac{C}{(x+1)^2}\\ 1 &= A(x+1)^2 + Bx(x+1) +Cx \end{align}\]

Now, to find the three unknowns, we'll take advantage of the fact that if these two sides of the equation are equal, they must be equal no matter what value \(x\) takes. We can therefore plug in different values of \(x\) and still have an equation; the reason this helps is that if we pick carefully what values of \(x\) to plug in, we can simplify things to the point where it's easy to find the unknowns.

For instance, if we plug in \(x=0\), all but one term on the right side will disappear, making it easy to find \(A\): \[1 = A(0+1)^2 + B(0)(0+1) + C(0) \longrightarrow 1 = A\] Along the same lines, we can let \(x=-1\), so that the factor \(x+1\) will be zero, making it easy to find \(C\): \[1 = A(-1+1)^2 + B(-1)(-1+1) + C(-1) \longrightarrow 1 = C(-1) \longrightarrow C = -1\]

When we try to find \(B\), there's no other value of \(x\) that'll work quite as nicely, since there's no other factor that will disappear with a carefully chosen \(x\). Therefore, we just have to pick any other value for \(x\) that we like, and use the fact that we already know what \(A\) and \(C\) are. For example, say we let \(x=1\): \[\begin{align} 1 &= A(1+1)^2 + B(1)(1+1) + C(1)\\ 1 &= 1(1+1)^2 + B(1)(1+1) + (-1)(1)\\ 1 &= 4 + 2B - 1\\ -2 &= 2B\\ -1 &= B \end{align}\]

We've found \(A\), \(B\), and \(C\), so now we've found that \[\dfrac{1}{x^3+2x^2+x} = \dfrac{1}{x(x+1)^2} = \dfrac{1}{x} + \dfrac{-1}{x+1} + \dfrac{-1}{(x+1)^2}.\]

Finally, all that's left is to see the guidelines that tell us how to set up the denominators of the partial fractions. This all depends on what the factored form of the original rational function looks like, and I'll cover three possibilities.

\(\bbox[border:2px solid black,8pt]{\text{Case } 1:}\) There are only linear factors (meaning they have the form \(ax+b\)), and none of the factors are repeated.

In this case, each factor gets its own partial fraction, and the numerators are constants. For example, \[\dfrac{3x}{x(x+4)(2x-1)} = \dfrac{A}{x} + \dfrac{B}{x+4} + \dfrac{C}{2x-1}\]

\(\bbox[border:2px solid black,8pt]{\text{Case } 2:}\) There are only linear factors and some of the factors are repeated (for instance, in the example above, the factor \(x+1\) was repeated twice).

In this case, each factor gets its own partial fraction, and each repeated factor gets as many terms as the number of times it is repeated (one for each power of that factor up to \(n\)). This is hard to explain, and easier to show. For example, \[\dfrac{1}{x^2(x+1)^3} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1} + \dfrac{D}{(x+1)^2} + \dfrac{E}{(x+1)^3}\]

\(\bbox[border:2px solid black,8pt]{\text{Case } 3:}\) There are some quadratic factors (meaning they have the form \(ax^2+bx+c\)), and these quadratic factors cannot be broken down into the product of linear factors.

In this case, the quadratic factor gets a partial fraction with a linear numerator, instead of a constant. For example, \[\dfrac{1}{x(x+1)(x^2+4x-9)} = \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{Cx+D}{x^2+4x-9}\]

Use partial fraction decomposition to write \(\dfrac{x^2+2x-1}{2x^3+3x^2-2x}\) as the sum of simpler rational functions.

Start by factoring the denominator: first, pull out the \(x\) that is common to all the terms, and then break down the remaining quadratic. \[2x^3+3x^2-2x = x(2x^2+3x-2) = x(2x-1)(x+2)\]

These are all linear factors, with no repetition, so by consulting the guidelines above, we can be confident that \[\dfrac{x^2+2x-1}{2x^3+3x^2-2x} = \dfrac{A}{x} + \dfrac{B}{2x-1} + \dfrac{C}{x+2}.\] To find the constants, multiply both sides by the denominator on the left, so that we can start plugging in \(x\) values that will simplify things: \[\begin{align} \dfrac{x^2+2x-1}{x(2x-1)(x+2)} &= \dfrac{A}{x} + \dfrac{B}{2x-1} + \dfrac{C}{x+2}\\ x(2x-1)(x+2) \cdot \dfrac{x^2+2x-1}{x(2x-1)(x+2)} &= x(2x-1)(x+2) \cdot \left(\dfrac{A}{x} + \dfrac{B}{2x-1} + \dfrac{C}{x+2}\right)\\ x^2+2x-1 &= A(2x-1)(x+2) + Bx(x+2) + Cx(2x-1) \end{align}\]

In order to make terms disappear, we should plug in \(x=0\), \(x=1/2\), and \(x=-2\), each of which will make one of the factors go to zero.

- If \(x=0\): \[\begin{align}0^2+2(0)-1 &= A(2(0)-1)((0)+2) + B(0)((0)+2) + C(0)(2(0)-1)\\ -1 &= A(-1)(2)\\ \dfrac{1}{2} &= A \end{align}\]
- If \(x=\dfrac{1}{2}\): \[\small{\begin{align} \left(\dfrac{1}{2}\right)^2+2\left(\dfrac{1}{2}\right)-1 &= A\left(2\left(\dfrac{1}{2}\right)-1\right)\left(\dfrac{1}{2}+2\right) + B\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}+2\right) + C\left(\dfrac{1}{2}\right)\left(2\left(\dfrac{1}{2}\right)-1\right)\\ \dfrac{1}{4} &= B\left(\dfrac{1}{2}\right)\left(\dfrac{5}{2}\right)\\ \dfrac{1}{4} &= B\left(\dfrac{5}{4}\right)\\ \dfrac{4}{5} \cdot \dfrac{1}{4} &= B\\ \dfrac{1}{5} &= B \end{align}}\]
- If \(x=-2\): \[\begin{align} (-2)^2+2(-2)-1 &= A(2(-2)-1)(-2+2) + B(-2)(-2+2) + C(-2)(2(-2)-1)\\ 4-4-1 &= C(-2)(-5)\\ -1 &= 10C\\ -\dfrac{1}{10} &= C \end{align}\]

Therefore, \[\dfrac{x^2+2x-1}{2x^3+3x^2-2x} = \ans{\dfrac{1/2}{x} + \dfrac{1/5}{2x-1} + \dfrac{-1/10}{x+2}}\]

Use partial fraction decomposition to integrate \(\displaystyle\int \dfrac{x}{x^2+x-2} \ dx.\)

Factoring this denominator is pretty straightforward: \[x^2+x-2 = (x+2)(x-1)\] Since these are both linear factors with no repetition, we'll have two partial fractions, like so: \[\dfrac{x}{x^2+x-2} = \dfrac{A}{x+2} + \dfrac{B}{x-1}\]

Again, to find \(A\) and \(B\), clear the denominators: \[x = A(x-1) + B(x+2).\] Now, plug in \(x=1\) and \(x=-2\):

- If \(x=1\): \[\begin{align}1 &= A(1-1) + B(1+2)\\ 1 &= B(3)\\ \dfrac{1}{3} &= B \end{align}\]
- If \(x=-2\): \[\begin{align}-2 &= A(-2-1) + B(-2+2)\\ -2 &= A(-3)\\ \dfrac{2}{3} &= A \end{align}\]

Therefore, \[\dfrac{x}{x^2+x-2} = \dfrac{2/3}{x+2} + \dfrac{1/3}{x-1},\] and we can integrate each of those pieces (again with a simple u-substitution that I won't show in detail: \[\int \dfrac{x}{x^2+x-2} \ dx = \int \dfrac{2/3}{x+2} + \dfrac{1/3}{x-1} \ dx = \ans{\dfrac{2}{3} \ln |x+2| + \dfrac{1}{3} \ln |x-1| + C}\]

For the sake of brevity, I often skip the last step in problems like these, going from something like \(\displaystyle\int \dfrac{2/3}{x+2} \ dx\) to \(\dfrac{2}{3}\ln |x+2|\). You should only do this after you've done enough similar integrals (by u-substitution) to ensure that you don't miss something simple. I'll warn you of a few mistakes that I often see students make:

- Ignoring coefficients of \(x\) in the denominator:
\[\int \dfrac{3}{1-2x} \ dx \neq 3 \ln |1-2x| + C\]\[\int \dfrac{3}{1-2x} \ dx = -\dfrac{3}{2} \ln |1-2x| + C\]
- Taking anything in the denominator and jumping to the natural log:
\[\int \dfrac{1}{(x+1)^2} \ dx \neq \ln |(x+1)^2| + C\]\[\int \dfrac{1}{(x+1)^2} \ dx = -\dfrac{1}{x+1} + C\]

I've purposely left these without explanation; you should do the u-substitution on your own to double-check them. Once you get familiar with these common pitfalls, and are able to successfully avoid them, you should be able to do the last step of a PFD problem just as easily.

\(\displaystyle\int \dfrac{x-9}{(x+5)(x-2)} \ dx\)

\(2 \ln |x+5| - \ln |x-2| + C\)

Evaluate the following integral: \(\displaystyle\int \dfrac{x^4-1}{x^5+4x^3} \ dx.\)

To factor the denominator, factor out \(x^3\) and notice that what remains is a quadratic expression that we cannot factor further: \[x^5+4x^3 = x^3(x^2+4)\] Be careful here; the \(x\) factor is simply repeated three times (it's not an irreduceable cubic), so we'll need three partial fractions for that one. \[\dfrac{x^4-1}{x^5+4x^3} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x^3} + \dfrac{Dx+E}{x^2+4}\] After we clear the denominators: \[x^4-1 = Ax^2(x^2+4) + Bx(x^2+4) + C(x^2+4) + (Dx+E)x^3.\] We can already tell that this is going to be a tough one, just because we have five unknowns to find. Never mind, once more into the breach!

Notice that the only value we can plug in that will make some factors disappear is \(x=0\). To use the method we've used so far, we'd have to pick four other points (small numbers to make the arithmetic doable), and the resulting system of equations would be tedious to solve (although you could have a calculator do it for you). Instead, I'll do the other method (outline in the footnotes) for this one, just to see an example of it.

Start by multiplying everything out on the right-hand side, then collect like terms. When you do this, you should get the following: \[x^4-1 = (A+D)x^4 + (B+E)x^3 + (4A+C)x^2 + 4Bx + 4C.\] Now comes the important part: all the terms on the left side should like up with all the terms on the right. Equating the coefficients, we get the following: \[\begin{align} A + D &= 1\\ B+E &= 0\\ 4A+C &= 0\\ 4B &= 0\\ 4C &= -1 \end{align}\]

The last two equations give us answers for \(B\) and \(C\), and taking those to the second and third equations gives us \(A\) and \(E\). Finally, the first equation then gives us \(D\). Summarizing these: \[\begin{align} A &= \dfrac{1}{16}\\ B &= 0\\ C &= -\dfrac{1}{4}\\ D &= \dfrac{15}{16}\\ E &= 0 \end{align}\]

Now we're finally ready to integrate these partial fractions. Notice that we have to do another u-substitution on the last term (details omitted; you should be able to fill them in). \[\begin{align} \int \dfrac{x^4-1}{x^5+4x^3} \ dx &= \int \dfrac{1/16}{x} - \dfrac{1/4}{x^3} + \dfrac{15x/16}{x^2+4} \ dx\\ \\ &= \ans{\dfrac{1}{16} \ln x + \dfrac{1}{8x^2} + \dfrac{15}{32} \ln |x^2+4| + C} \end{align}\]

\(\displaystyle\int \dfrac{5x^2+3x-2}{x^3+2x^2} \ dx\)

\(2 \ln x +\dfrac{1}{x} + 3 \ln |x+2| + C\)

Evaluate the following integral: \(\displaystyle\int \dfrac{x^2+2x-1}{x^3-x} \ dx.\)

Letting \(x=0, 1,\) and \(-1\):

- If \(x=0\): \[-1 = A(1)(-1) \longrightarrow A = 1\]
- If \(x=1\): \[2 = C(1)(2) \longrightarrow C = 1\]
- If \(x=-1\): \[-2 = B(-1)(-2) \longrightarrow B = -1\]

- \(\displaystyle\int \dfrac{x}{x^2+x-2} \ dx\)
- \(\displaystyle\int \dfrac{x^2-5x+16}{(2x+1)(x-2)^2} \ dx\)

\(\dfrac{2}{3} \ln |x+2| + \dfrac{1}{3} \ln |x-1| + C\)

\(\dfrac{3}{2} \ln |2x+1| - \ln |x-2| - \dfrac{2}{x-2} + C\)