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Section 7.7 Partial Fractions

In this last integration technique, we’ll consider more rational functions. We’ve already thought about rational functions a bit (Integrating Rational Functions), but here we’ll add some more detail to a special type of rational function. Let’s not spoil anything. Instead, we’ll just jump into a quick comparison.

Activity 7.7.1 Comparing Rational Integrands.

We’re going to compare three integrals:
\begin{equation*} \int \frac{2}{x^2+4x+5}\;dx \end{equation*}
\begin{equation*} \int \frac{2}{x^2+4x+3}\;dx \end{equation*}
\begin{equation*} \int \left(\frac{1}{x+1} - \frac{1}{x+3}\right)\;dx \end{equation*}

(a)

Start with the first integral:
\begin{equation*} \int \frac{2}{x^2+4x+5}\;dx\text{.} \end{equation*}
How would you approach integrating this?
Hint.
This is a constant over a quadratic: can you complete the square, and then connect the result to an inverse tangent function?

(b)

Try the same tactic on the second integral:
\begin{equation*} \int \frac{2}{x^2+4x+3}\;dx\text{.} \end{equation*}
You don’t need to complete this integral, but think about how you might proceed.
Hint.
Check the structure of the denominator when you complete the square: since you don’t end up with a sum of squares, you can’t use the inverse tangent setup. You’ll need to do a trig substitution instead...

(c)

Think about the third integral:
\begin{equation*} \int \left(\frac{1}{x+1} - \frac{1}{x+3}\right)\;dx\text{.} \end{equation*}
How would you integrate this?

(d)

The third integral is unique from the other two in that it is has two terms. Let’s combine them together to see how we could write this integral to compare it more closely to the other two.
Subtract \(\dfrac{1}{x+1}-\dfrac{1}{x+3}\) using common denominators and compare your rewritten integral to the other two.

(e)

Which of these integrals and/or representations of an integral is easiest to work with? Which one is most annoying to work with? Why?
We’re going to try to take advantage of the rewritten structure in Activity 7.7.1: when we can decompose a “large” rational function into a sum of “smaller” rational functions, we’ll be more likely to be able to antidifferentiate the “smaller” pieces.
The real trick, here, is going to be recognizing when we can do this and building a process for how we do this.

Subsection When?

In order to recognize when we will employ this strategy, we should think about what we’re doing: we are attempting to write one fraction as a sum or difference of others. We’re “undo-ing” fraction addition, here. We can remember that when we add (or subtract) fractions, we need to find a common denominator and rewrite our fraction terms as equivalent versions with this same denominator. This typically is done by just “scaling” the numerator and denominator of a fraction by a factor (often the other denominator). For instance:
\begin{align*} \frac{3}{5} - \frac{1}{3} \amp = \frac{3}{5}\left(\frac{3}{3}\right) - \frac{1}{3}\left(\frac{5}{5}\right)\\ \amp = \frac{9}{15} - \frac{5}{15}\\ \amp = \frac{4}{15} \end{align*}
This same thing happens when we think about rational functions:
\begin{align*} \frac{1}{x} + \frac{2}{x-1} \amp = \frac{1}{x}\left(\frac{x-1}{x-1}\right) + \frac{2}{x-1}\left(\frac{x}{x}\right)\\ \amp = \frac{x-1}{x(x-1)} + \frac{2x}{x(x-1)}\\ \amp =\frac{3x-1}{x(x-1)}\;\;\text{ or } \;\;\frac{3x-1}{x^2-x} \end{align*}
Now, we can notice something about these (common) denominators: since we scaled each fraction before adding/subtracting them, the denominator is a product of factors.
So right away we know that this method only will work well when we can factor the denominator of a rational integrand.
We also know (from Integrating Rational Functions) that once the degree of the numerator is at least the same (or larger) than the degree of the denominator, we can rewrite things using division.
So, we will use this strategy of rewriting rational function integrands as sums or differences of “smaller” rational functions when:
  1. The denominator can be factored.
  2. The degree of the denominator is larger than the degree of the numerator.

Subsection How?

The actual process for finding the smaller “partial” fractions is not complicated. Once we have the denominator of our rational function factored, we can see what the possible denominators that had to “combine” to make a “common” denominator were.
Our general process will be simple: set up these possible rational functions and put unknown placeholder numerators in place, making sure that the degree is smaller than the degree of the denominator. Then, we add these possible rational functions up and compare the numerator (with the unknown placeholders) to the numerator that we want (from our actual function that we’re integrating).
There are more tricks along the way, but this process is simple to think about. What we’ll find, though, is that the process is full of algebra, which can be tedious.
Let’s look at two small examples to see how this could work.

Activity 7.7.2 First Examples of Partial Fractions.

(a)

Consider the integral:
\begin{equation*} \int \frac{6x-16}{x^2-4x+3}\;dx\text{.} \end{equation*}
First, confirm that this would be very annoying to try to use \(u\)-substitution on, even though we have a linear numerator and quadratic denominator.

(b)

Notice that the denominator can be factored:
\begin{equation*} \int \frac{6x-16}{(x-3)(x-1)}\;dx\text{.} \end{equation*}
If this integrand function were a sum of two “smaller” rational functions, what would their denominators be? What do we know about their numerators?
Hint.
The denominators would need to multiply to get \((x-3)(x-1)\text{,}\) and each numerator would have to have a degree smaller than its respective denominator.
Solution.
We might have something like \(\dfrac{\fillinmath{XX}}{x-3}+\dfrac{\fillinmath{XX}}{x-1} = \dfrac{6x-16}{(x-3)(x-1)}\text{,}\) where the numerators are constant terms (since their degree must be smaller than 1).

(c)

Use some variables (it’s typical to use capital letters like \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) etc.) to represent the numerators, and then add the partial fractions together. What do you get? How does this rational function compare to \(\dfrac{6x-16}{(x-3)(x-1)}\text{?}\)
Hint.
We want to try to add up \(\dfrac{A}{x-3}+\dfrac{B}{x-1}\) and compare the result to \(\dfrac{6x-16}{(x-3)(x-1)}\text{.}\)
Solution.
\begin{align*} \frac{A}{x-3} + \frac{B}{x-1} \amp = \frac{A}{x-3}\left(\frac{x-1}{x-1}\right) + \frac{B}{x-1}\left(\frac{x-3}{x-3}\right)\\ \amp = \frac{A(x-1)}{(x-3)(x-1)} + \frac{B(x-3)}{(x-3)(x-1)}\\ \amp = \frac{A(x-1)+B(x-3)}{(x-3)(x-1)} \end{align*}
When we compare this to \(\dfrac{6x-16}{(x-3)(x-1)}\text{,}\) we can see that the denominators are the same, and so, if these are equal to each other, the numerators must be as well.

(d)

Set up an equation connecting the numerators, and solve for your unknown variables. What are the two fractions that added together to get \(\dfrac{6x-16}{(x-3)(x-1)}\text{?}\)
Solution.
\begin{align*} A(x-1)+B(x-3)\amp = 6x-16\\ (A+B)x -(A+3B) \amp = 6x-16 \end{align*}
This means that \(A+B=6\) and \(A+3B=16\text{.}\) There are a variety of ways to solve this, but we can say that \(A=6-B\text{,}\) and so then
\begin{align*} A+3B \amp=16 \\ (6-B)+3B \amp = 16\\ 2B \amp = 10\\ B \amp =5 \end{align*}
Since \(A=6-B\text{,}\) we have \(A=6-5=1\text{.}\)
\begin{equation*} \dfrac{6x-16}{(x-3)(x-1)} = \frac{1}{x-3}+\frac{5}{x-1} \end{equation*}

(e)

Antidifferentiate to solve the integral \(\displaystyle \int \frac{6x-16}{(x-3)(x-1)}\;dx\text{.}\)
Solution.
\begin{align*} \int \frac{6x-16}{(x-3)(x-1)}\;dx\amp = \int \frac{1}{x-3}+\frac{5}{x-1}\;dx\\ \amp =\ln|x-3|+5\ln|x-1|+C \end{align*}

(f)

Let’s do the same thing with a new integral:
\begin{equation*} \int \frac{3x^2-2x+3}{x(x^2+1)}\;dx\text{.} \end{equation*}
What are the partial fraction forms? What kinds of numerators should we expect to see? Use variables to represent these.
Solution.
We expect to see partial fractions \(\dfrac{\fillinmath{XX}}{x}\) and \(\dfrac{\fillinmath{XX}}{x^2+1}\text{.}\) We know the numerator over \(x\) needs to be constant, but we could have a linear numerator over \(x^2+1\text{.}\) So our partial fractions are:
\begin{equation*} \frac{A}{x}+\frac{Bx+C}{x^2+1}\text{.} \end{equation*}

(g)

Add the partial fractions together and set up an equation for the numerators to solve. What are the two partial fractions after you solve for the unknown coefficients?
Hint.
We want to find values for \(A\text{,}\) \(B\text{,}\) and \(C\) where
\begin{equation*} \frac{A}{x}+\frac{Bx+C}{x^2+1} = \frac{3x^2-2x+3}{x(x^2+1)}\text{.} \end{equation*}
Use common denominators to add
\begin{equation*} \frac{A}{x}+\frac{Bx+C}{x^2+1} \end{equation*}
and then set the resulting numerator equal to \(3x^2-2x+3\) and try to solve.
Solution.
\begin{align*} \frac{A}{x}+\frac{Bx+C}{x^2+1}\amp = \frac{A}{x}\left(\frac{x^2+1}{x^2+1}\right)+\frac{Bx+C}{x^2+1}\left(\frac{x}{x}\right) \\ \amp = \frac{A(x^2+1)+(Bx+C)(x)}{x(x^2+1)} \end{align*}
Then, we can set the numerators equal to get:
\begin{align*} A(x^2+1)+(Bx+C)(x)\amp = 3x^2-2x+3\\ (A+B)x^2 + Cx + A \amp = 3x^2-2x+3 \end{align*}
So, we now can see that since \(Cx=-2x\text{,}\) we have \(C=-2\text{.}\) We also can look at the constant terms and see that \(A=3\text{.}\) Now we know that \((A+B)x^2=3x^2\text{,}\) but since \(A=3\) we have \(B=0\text{.}\)
So we have \(\displaystyle \int \frac{3x^2-2x+3}{x(x^2+1)}\;dx = \int \frac{3}{x}-\frac{2}{x^2+1}\;dx\text{.}\)

(h)

Antidifferentiate and solve the integral \(\displaystyle \int \frac{3x^2-2x+3}{x(x^2+1)}\;dx\text{.}\)
Solution.
\begin{align*} \int \frac{3x^2-2x+3}{x(x^2+1)}\;dx \amp = \int \frac{3}{x}-\frac{2}{x^2+1}\;dx\\ \amp = 3\ln|x|-2\tan^{-1}(x)+C \end{align*}
So we can see the basics of how this will work:
  1. Figure out the denominators of the fractions we could add to get the function we’re integrating.
  2. Construct the partial fractions using placeholders for the numerators (making sure to keep the degree of the numerators smaller than the degree of the denominators).
  3. Add these placeholder fractions up, and see what the coefficients would have to be in order to make them add to the function we’re integrating.
  4. Antidifferentiate the smaller rational functions.

Subsection More Specific Strategies

We’re going to investigate these partial fractions a bit more, and focus on two cases: linear denominators, and quadratic denominators. This will limit the scope of our work enough that this doesn’t get too wild, and also works well with the kinds of things we know how to antidifferentiate: we can antidifferentiate any rational function with a linear denominator and we know how to antidifferentiate many rational functions with quadratic denominators.

Subsubsection Partial Fraction Type: Simple Linear Factors

For rational functions where the denominator has some linear factor like \((x-k)\text{,}\) we can set up a partial fraction with just a constant coefficient in the numerator:
\begin{equation*} \frac{A}{x-k}\text{.} \end{equation*}
We have outlined a pretty reasonable approach for these in Activity 7.7.2 when we worked on the integral
\begin{equation*} \int \frac{6x-16}{(x-3)(x-1)}\;dx\text{.} \end{equation*}
We can make note of something useful, though. At one point during the process, we had set up the two partial fractions, added them together, and then we said that we wanted to find the values of \(A\) and \(B\) that made the general partial fraction forms we had set up match with the actual function we were hoping to integrate.
\begin{align*} \frac{A}{x-3} + \frac{B}{x-1} \amp = \frac{A(x-1) + B(x-3)}{(x-3)(x-1)}\\ \amp = \frac{6x-16}{(x-3)(x-1)} \end{align*}
What this meant for us was that we can set the numerators equal to each other and solve for \(A\) and \(B\text{:}\)
\begin{equation*} A(x-1)+B(x-3)=6x-16\text{.} \end{equation*}
Let’s pause here.
We are hoping to find the values of \(A\) and \(B\) that will make this equation work out for all values of \(x\text{.}\) But we also can make use of the fact that this equation will be true for some specific values of \(x\text{.}\) For instance, we can evaluate this equation at \(x=2\text{:}\)
\begin{align*} A(2-1)+B(2-3) \amp = 6(2)-16\\ A-B \amp = -4 \end{align*}
This just generates another equation connecting \(A\) and \(B\) that we could use like we did in in Activity 7.7.2.
But is there a nicer, more convenient \(x\)-value to use? Can we force this to generate a nicer equation involving our unknown coefficients? Can we find some \(x\)-values that will make the factors \((x-1)\) and \((x-3)\) go to 0, for instance?
\begin{equation*} A\underbrace{(x-1)}_{x=1}+B\underbrace{(x-3)}_{x=3}=6x-16 \end{equation*}
Let’s try it!
\begin{align*} x=1: \amp \amp A(1-1)+B(1-3) \amp = 6(1)-16 \amp\\ \amp \amp -2B \amp= -10 \amp\\ \amp \amp B \amp = 5\\ x=3: \amp \amp A(3-1)+B(3-3)\amp = 6(3)-16\\ \amp \amp 2A \amp = 2\\ \amp \amp A \amp = 1 \end{align*}
These are the same values we had in Activity 7.7.2! This strategy works well for any simple linear factors, and definitely helps to reduce the amount of algebra required.

Simple Linear Factors.

If a rational function has a simple linear factor in the denominator in the form \((x-k)\text{,}\) then the corresponding partial fraction is
\begin{equation*} \frac{A}{x-k} \end{equation*}
where \(A\) is a real number constant with \(A\neq 0\text{.}\)
When we antidifferentiate these, we will end up with a logarithmic function.

Subsubsection Partial Fraction Type: Irreducible Quadratic Factors

So what about our other example in Activity 7.7.2? We had a quadratic factor (that couldn’t be factored nicely). We’ll call these irreducible quadratic factors. The real difference just was the structure of the numerator, where we accounted for the option of the numerator being some linear function, \(Ax+B\text{.}\)
There isn’t much more to say here, since the algebra can be frustrating to deal with. It can be helpful to save some of these coefficients for last: that way, we can find some of the “easier” ones (from the Simple Linear Factors cases) first, and hopefully the remaining coefficients won’t be too difficult to find.

Irreducible Quadratic Factors.

If a rational function has an irreducible quadratic factor in the denominator in the form \((ax^2+bx+c)\text{,}\) then the corresponding partial fraction is
\begin{equation*} \frac{Ax+B}{ax^2+bx+c} \end{equation*}
where \(A\) and \(B\) are real number constants, and \(A\) and \(B\) cannot both be \(0\text{.}\) That is, if one of \(A\) or \(B\) is \(0\text{,}\) then the other cannot be.
When we antidifferentiate these, we can expect a logarithmic function, an inverse tangent function, or some combination of the two. We can get other antiderivatives, but those will be beyond the scope of this introductory text.

Subsubsection Partial Fraction Type: Repeated Linear Factors

Let’s look at one last case. It’s a bit of a weird one, so we’ll explore it in its own example.

Activity 7.7.3 Fiddling with Repeated Factors.

Let’s sit with the following integral:
\begin{equation*} \int \frac{3x+5}{x^2+2x+1}\;dx\text{.} \end{equation*}
Before we start, we can think about how annoying it would be to try to start with a \(u\)-substitution where \(u=x^2+2x+1\text{.}\)
(a)
Factor the denominator! What’s different about the factors in this denominator compared to the ones in Activity 7.7.2?
Answer.
\(x^2+2x+1=(x+1)(x+1)\)
(c)
Ok, instead, we’re going to do some algebra that is reminiscent of what we have done before in Section 7.3.
Can you write the numerator, \(3x+5\text{,}\) as some coefficient on the factor \((x+1)\) with some constant “remainder?”
\begin{equation*} 3x+5 = \fillinmath{XX}(x+1)+\fillinmath{XX} \end{equation*}
Answer.
\begin{equation*} 3x+5 = 3(x+1)+2 \end{equation*}
(d)
Why is this reforming of the numerator useful? What does that do, when we write it over the factored denominator? Why did we choose \((x+1)\) as the factor to use for our rewriting?
Feel free to show why this is helpful!
Hint.
Split the fraction up across the sum in the numerator!
Solution.
\begin{align*} \frac{3x+5}{(x+1)^2} \amp = \frac{3(x+1)+2}{(x+1)^2}\\ \amp = \frac{3(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2}\\ \amp = \frac{3}{x+1}+\frac{2}{(x+1)^2} \end{align*}
(e)
Integrate \(\displaystyle\int\frac{3x+5}{x^2+2x+1}\;dx\) using your clever rewriting. Explain why this is a friendlier form.
Solution.
We’ll use our rewritten integral:
\begin{equation*} \int \frac{3x+5}{x^2+2x+1}\;dx = \int \frac{3}{x+1}+\frac{2}{(x+1)^2}\;dx\text{.} \end{equation*}
Now, we can split this into two integrals to deal with separately. For both, we can use the \(u\)-substitution \(u=x+1\text{.}\)
\begin{align*} \int \frac{3}{x+1}+\frac{2}{(x+1)^2}\;dx \amp = \int \frac{3}{u}\;du + \int \frac{2}{u^2}\;du\\ \amp = 3\ln|u| - \frac{2}{u}+C\\ \amp = 3\ln|x+1| - \frac{2}{x+1}+C \end{align*}
This is something we can do, algebraically, for every fraction with a “repeated” factor like this. But, more importantly, we can incorporate this idea into how we think about partial fractions.

Repeated Linear Factors.

If a rational function has a repeated linear factor in the denominator in the form \((x-k)^n\text{,}\) where m\(n\) is some integer greater than 1, then the corresponding partial fractions are
\begin{equation*} \frac{A_1}{x-k} + \frac{A_2}{(x-k)^2} + ... + \frac{A_n}{(x-k)^n} \end{equation*}
where \(A_1, A_2, ..., A_n\) are real numbers and \(A_n\neq 0\text{.}\)
When we antidifferentiate these, we can expect to use the rule and maybe find a logarithm function.
At this point, we can spend our time going through one or two examples where we put this all together.

Example 7.7.4.

For each of the following integrals, set up the relevant partial fraction forms, solve for the unknown coefficients, and then antidifferentiate.

(a)

\(\displaystyle \int \frac{5x^2+27x+51}{(x-2)(x+3)^3}\;dx\)
Hint 1.
Your partial fraction forms will look like this:
\begin{equation*} \frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^2}+\frac{D}{(x+3)^3}\text{.} \end{equation*}
Hint 2.
You can find the values for \(A\) and \(D\) pretty easily by thinking about convenient \(x\)-values, like we talked about in Partial Fraction Type: Simple Linear Factors.
Solution.
We’ll rewrite our integral using the partial fraction forms:
\begin{equation*} \int \frac{5x^2+27x+51}{(x-2)(x+3)^3}\;dx = \int \frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^2}+\frac{D}{(x+3)^3}\;dx\text{.} \end{equation*}
When we combine these fractions to compare the numerators, we end up with the following equation:
\begin{equation*} A(x+3)^3 + B(x-2)(x+3)^2 + C(x-2)(x+3)+D(x-2) = 5x^2+27x+51 \end{equation*}
We can evaluate this at \(x=2\) and \(x=-3\) to reveal the values of \(A\) and \(D\text{:}\)
\begin{align*} x=2: \amp \amp A(5)^3 \amp = 5(4)+27(2)+51 \amp\\ \amp \amp A \amp =\frac{125}{125}=1\\ x=-3: \amp \amp D(-5) \amp = 5(9)+27(-3)+51\\ \amp \amp D \amp = \frac{15}{-5}=-3 \end{align*}
Now that we know \(A=1\) and \(D=-3\text{,}\) we can put those into our equation connecting the numerators, and solve for \(B\) and \(C\text{.}\)
\begin{equation*} (x+3)^3 + B(x-2)(x+3)^2 + C(x-2)(x+3)-3(x-2) = 5x^2+27x+51 \end{equation*}
If we just consider the cubic terms, then on the left side of the equation we have \(x^3+Bx^3\text{,}\) and there are no cubic terms on the right side. This means that \(1+B=0\) and so \(B=-1\text{.}\)
Similarly, we can consider just the constant terms of the (updated) equation:
\begin{equation*} (x+3)^3 - (x-2)(x+3)^2 + C(x-2)(x+3)-3(x-2) = 5x^2+27x+51 \end{equation*}
We can see that on the left side, we’ll have \((3)^3-(-2)(3)^2 + C(-2)(3) - 3(-2)\) and on the right side, the constant term is \(51\text{.}\)
\begin{align*} 27+18-6C+6 \amp = 51\\ 51 - 6C \amp = 51\\ C \amp =0 \end{align*}
Finally, we have our new, rewritten, integral. We can antidifferentiate.
\begin{align*} \int \frac{5x^2+27x+51}{(x-2)(x+3)^3}\;dx \amp= \int \frac{1}{x-2}-\frac{1}{x+3} - \frac{3}{(x+3)^3}\;dx \\ \amp = \ln|x-2|-\ln|x+3| + \frac{3}{2(x+3)^2}+C \end{align*}

(b)

\(\displaystyle \int \frac{6x^2+25x+27}{(x+1)(x^2+4x+5)}\;dx\)
Hint 1.
Your partial fraction forms will look like this:
\begin{equation*} \frac{A}{x+1} + \frac{Bx+C}{x^2+4x+5}\text{.} \end{equation*}
Hint 2.
You’ll be able to easily find \(A\) by thinking about convenient \(x\)-values, but not \(B\) or \(C\text{.}\)
Solution.
Let’s, again, rewrite our integral using the partial fraction forms we set up:
\begin{equation*} \int \frac{6x^2+25x+27}{(x+1)(x^2+4x+5)}\;dx = \int frac{A}{x+1} + \frac{Bx+C}{x^2+4x+5}\;dx \end{equation*}
Our equation for the combined numerator is:
\begin{equation*} A(x^2+4x+5) + (Bx+C)(x+1) = 6x^2+25x+27\text{.} \end{equation*}
We can find \(A\) be evaluating at \(x=-1\text{.}\)
\begin{align*} x=-1:\amp \amp A(1-4+5) \amp = 6(1)-25+27\amp\\ \amp \amp A\amp= \frac{8}{2} = 4 \end{align*}
Now, knowing that \(A=2\text{,}\) we can rewrite our equation to solve for \(B\) and \(C\text{.}\)
\begin{equation*} 4(x^2+4x+5) + (Bx+C)(x+1) = 6x^2+25x+27 \end{equation*}
We can collect the quadratic terms, and see the following equation:
\begin{equation*} 4x^2 + Bx^2 = 6x^2 \end{equation*}
So \(B=2\text{.}\)
Similarly, we can collect the constant terms:
\begin{equation*} 4(5) + C = 27\text{.} \end{equation*}
It is easy to see that \(C=7\text{.}\) So we have our newly rewritten integral:
\begin{equation*} \int \frac{6x^2+25x+27}{(x+1)(x^2+4x+5)}\;dx = \int \frac{4}{x+1}+ \frac{2x+7}{x^2+4x+5}\;dx \end{equation*}
The first term is pretty straightforward to integrate: we’ll get a log. The second one, though, will take some work. Let’s consider it by itself:
\begin{equation*} \int \frac{2x+7}{x^2+4x+5}\;dx\text{.} \end{equation*}
We can start with a \(u\)-substitution of \(u=x^2+4x+5\text{,}\) giving us \(dx=2x+4\;dx\text{.}\) Let’s rewrite the numerator as \(2x+4+3\) in order to make this work:
\begin{align*} \int \frac{2x+7}{x^2+4x+5}\;dx \amp =\int \frac{(2x+4)+3}{x^2+4x+5}\;dx\\ \amp = \int \frac{2x+4}{x^2+4x+5}\;dx + \int \frac{3}{x^2+4x+5}\;dx \end{align*}
Now, the first of these will work with our stated substitution. The second one, though, will require a different strategy. Let’s complete the square to get the inverse tangent form (Theorem 7.3.6). For all three integrals, then, we get:
\begin{align*} \int \frac{6x^2+25x+27}{(x+1)(x^2+4x+5)}\;dx \amp = \int \frac{4}{x+1}+ \frac{2x+7}{x^2+4x+5}\;dx\\ \amp = \int \frac{4}{x+1}+ \frac{2x+4}{x^2+4x+5} + \frac{3}{x^2+4x+5}\;dx\\ \amp =\int \frac{4}{x+1}+\frac{2x+4}{x^2+4x+5} + \frac{3}{(x+2)^2+1}\;dx\\ \amp = 4\ln|x+1| + \ln|x^2+4x+5| + 3\tan^{-1}(x+2)+C \end{align*}
There are more things that we can think about, but it really ends up being just extensions of what we’ve done. For instance, we could think about repeated quadratic factors or irreducible polynomials that have larger degrees, but the general principles are the same: we set up a placeholder numerator that has a degree less than the denominator and try to solve for the unknown coefficients.
There really are only two limitations for us:
  1. As we increase the number of coefficients, it becomes very tedious to solve for them. It isn’t difficult, really: just a lot of algebra.
  2. As we increase the degree of the kinds of denominators we see, we run out of approaches for antidifferentiation. We could spend much more time talking about integrating more rational functions or dive into the world of irrational coefficients (or even non-real ones), but this serves as a good stopping point for our purposes.

Practice Problems Practice Problems

1.

Why do we use partial fraction decomposition on some integrals of rational functions? Give an example and explain why it is helpful in your example.

2.

For each rational function described, write out the corresponding partial fraction forms.

(a)

\(\dfrac{p(x)}{(x-4)(x+2)(x-1)}\) where \(p(x)\) is some polynomial with degree less than 3.
Solution.
\begin{equation*} \dfrac{p(x)}{(x-4)(x+2)(x-1)} = \frac{A}{x-4} + \frac{B}{x+2} + \frac{C}{x-1} \end{equation*}

(b)

\(\dfrac{p(x)}{(x+1)^2(3x-5)^3}\) where \(p(x)\) is some polynomial with degree less than 5.
Solution.
\begin{equation*} \dfrac{p(x)}{(x+1)^2(3x-5)^3} = \frac{A}{(x+1)^2} + \frac{B}{x+1} + \frac{C}{(3x-5)^3} + \frac{D}{(3x-5)^2} + \frac{E}{3x-5} \end{equation*}

(c)

\(\dfrac{p(x)}{(x^2+1)(x^2+2x+5)}\) where \(p(x)\) is some polynomial with degree less than 4.
Solution.
\begin{equation*} \dfrac{p(x)}{(x^2+1)(x^2+2x+5)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2x+5} \end{equation*}

(d)

\(\dfrac{p(x)}{x^4-1}\) where \(p(x)\) is some polynomial with degree less than 4.
Hint.
There’s some factoring to be done here! Note that \(x^4-1 = (x^2-1)(x^2+1)\) and then we can factor \(x^2-1 = (x-1)(x+1)\text{.}\)
Solution.
\begin{equation*} \frac{p(x)}{x^4-1} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1} + \frac{D}{x+1} \end{equation*}

3.

Consider the following integral, with the partial fraction forms written out:
\begin{equation*} \int \frac{x^3+6x^2-x}{(x-2)(x+1)(x^2+1)}\;dx = \int \left(\frac{A}{x-2} + \frac{B}{x+1}+\frac{Cx+D}{x^2+1}\right)\;dx\text{.} \end{equation*}

(a)

Write an equation connecting the numerators.
Solution.
\begin{equation*} A(x+1)(x^2+1) + B(x-2)(x^2+1) + (Cx+D)(x-2)(x+1) = x^3+6x^2-x \end{equation*}

(b)

Find (and use) a specific \(x\)-value to input into the equation to solve for \(A\text{.}\)
Hint.
Use \(x=2\text{,}\) and notice what happens to the rest of the terms.
Solution.
\begin{align*} A(2+1)(2^2+1) + B(2-2)(2^2+1) + (C2+D)(2-2)(2+1) \amp = 2^3+6(2^2)-2 \\ 3(5)A \amp = 30\\ A \amp = 2 \end{align*}

(c)

Find (and use) a specific \(x\)-value to input into the equation to solve for \(B\text{.}\)
Hint.
Use \(x=-1\text{,}\) and notice what happens to the rest of the terms.
Solution.
\begin{align*} A(-1+1)((-1)^2+1) + B(-1-2)((-1)^2+1) + (C(-1)+D)(-1-2)(-1+1) \amp = (-1)^3+6(-1)^2-(-1)\\ -3(2)B \amp = 6\\ B \amp = -1 \end{align*}

(d)

Why can you not use this strategy to solve for coefficients \(C\) or \(D\text{?}\)

(e)

Find the cubic terms (you will need to do some multiplication) on both sides of your equation. Use these to solve for \(C\text{.}\)
Solution.
\begin{equation*} A(x+1)(x^2+1) + B(x-2)(x^2+1) + (Cx+D)(x-2)(x+1) = x^3+6x^2-x \end{equation*}
The cubic terms of this are:
\begin{align*} Ax^3 + Bx^3 + Cx^3 \amp x^3\\ A + B + C \amp = 1 \end{align*}
We already have \(A=2\) and \(B=-1\text{,}\) so we can solve for \(C\text{:}\)
\begin{align*} 2 + (-1) + C \amp = 1\\ C \amp = 0 \end{align*}

(f)

Find the constant terms (you will need to do some multiplication) on both sides of your equation. Use these to solve for \(D\text{.}\)
Solution.
\begin{equation*} A(x+1)(x^2+1) + B(x-2)(x^2+1) + (Cx+D)(x-2)(x+1) = x^3+6x^2-x \end{equation*}
The constant terms of this are:
\begin{align*} A -2 B -2D \amp = 0 \end{align*}
Since \(A=2\) and \(B=-1\text{,}\) we can solve for \(D\text{:}\)
\begin{align*} 2 - 2(-1) - 2D \amp = 0\\ 2D \amp = 4\\ D \amp = 2 \end{align*}

(g)

Integrate!
Solution.
\begin{align*} \int \frac{x^3+6x^2-x}{(x-2)(x+1)(x^2+1)}\;dx \amp = \int \left(\frac{A}{x-2} + \frac{B}{x+1}+\frac{Cx+D}{x^2+1}\right)\;dx\\ \amp = \int \left(\frac{2}{x-2} - \frac{1}{x+1}+\frac{2}{x^2+1}\right)\;dx\\ \amp = 2\ln|x-2| - \ln|x+1| + 2\tan^{-1}(x) + C \end{align*}

4.

Explain why partial fractions is not an appropriate technique for the following integral:
\begin{equation*} \int \dfrac{x^2+x}{x^2-x+1}\;dx\text{.} \end{equation*}
How should we approach this integral, instead? Use this to practice some other strategies for integrating rational functions!
Hint.
Note the degree in the numerator compared to the denominator!
Solution.
\begin{align*} \int \frac{x^2+x}{x^2-x+1}\;dx \amp = \int 1 + \frac{2x - 1}{x^2-x+1} \;dx\\ \amp = \int 1 \;dx + \int \frac{2x-1}{x^2-x+1}\;dx\\ \amp = x + \ln|x^2-x+1| + C \end{align*}

5.

Integrate the following.

(a)

\(\dint \dfrac{2}{(x-1)(x+3)}\;dx\)
Solution.
\begin{align*} int \dfrac{2}{(x-1)(x+3)}\;dx \amp = \int \frac{A}{x-1} + \frac{B}{x+3}\;dx\\ \amp = \int \frac{1/2}{x-1} - \frac{1/2}{x+3} \;dx\\ \amp = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+3| + C \end{align*}

(b)

\(\dint \dfrac{4x+1}{(x-4)(x+5)}\;dx\)
Solution.
\begin{align*} \int \frac{4x+1}{x-4}{x+5} \amp = \int \frac{A}{x-4} + \frac{B}{x+5}\;dx\\ \amp = \int \frac{17/9}{x-4} - \frac{21/9}{x+5}\;dx\\ \amp = \frac{17}{9}\ln|x-4| - \frac{21}{9}\ln|x+5| + C \end{align*}

(c)

\(\dint \dfrac{2x^2-15x+32}{x(x^2-8x+64)}\;dx\)
Solution.
\begin{align*} \int \dfrac{2x^2-15x+32}{x(x^2-8x+64)}\;dx \amp = \int \frac{A}{x} + \frac{Bx+C}{x^2-8x+64}\;dx \\ \amp = \int \frac{1/2}{x} + \frac{3x/2 - 11}{x^2-8x+64} \; dx\\ \amp = \frac{1}{2}\ln|x| + \frac{3}{2}\int \frac{x- 22/3}{x^2-8x+64}\;dx \end{align*}
In order to perform a \(u\)-substitution with \(u=x^2-8x+64\text{,}\) we’ll need \(du=2x-4\;dx\) or \(du = 2(x-2)\;dx\text{.}\) Let’s isolate \(x-2\) in the numerator and split the fraction.
\begin{align*} \int \frac{x- 22/3}{x^2-8x+64}\;dx \amp = \int \frac{x-2 - 16/3}{x^2-8x+64}\;dx\\ \amp = \int \frac{x-2}{x^2-8x+64} - \frac{16/3}{x^2-8x+64}\;dx\\ \amp = \frac{1}{2}\int \frac{1}{u}\;du - \int \frac{16/3}{(x-4)^2 + 48}\;dx\\ \amp = \frac{1}{2}\ln|x^2-8x+64| - \frac{4}{3\sqrt{3}}\tan^{-1}\left(\frac{x-4}{4\sqrt{3}}\right) + C \end{align*}
So, we end up with:
\begin{equation*} \frac{1}{2}\ln|x| + \frac{3}{2}\int \frac{x- 22/3}{x^2-8x+64}\;dx = \frac{1}{2}\ln|x| + \frac{3}{4}\ln|x^2-8x+64| - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x-4}{4\sqrt{3}}\right) + C \end{equation*}

(d)

\(\dint \dfrac{1}{(x+2)(x-2)}\;dx\)
Solution.
\begin{align*} \int \frac{1}{(x+2)(x-2)}\;dx \amp = \int \frac{A}{x+2} + \frac{B}{x-2}\;dx\\ \amp = \int -\frac{1/4}{x+2} + \frac{1/4}{x-2}\;dx\\ \amp = \frac{1}{4}\ln|x-2| - \frac{1}{4}\ln|x+2| + C \end{align*}

(e)

\(\dint \dfrac{20x}{(x-1)(x^2+4x+5)}\;dx\)
Solution.
\begin{align*} \int \dfrac{20x}{(x-1)(x^2+4x+5)}\;dx \amp = \int \frac{A}{x-1} + \frac{Bx+C}{x^2+4x+5}\;dx\\ \amp = \int \frac{2}{x-1} - \frac{2x-10}{x^2+4x+5}\;dx\\ \amp = 2\ln|x-1| - \int \frac{2x-10}{x^2+4x+5}\;dx \end{align*}
In order for us to use the substitution \(u=x^2+4x+5\text{,}\) we’ll need \(du=2x+4\;dx\text{,}\) so we should isolate \(2x+4\) in the numerator and split the fraction.
\begin{align*} \int \frac{2x-10}{x^2+4x+5}\;dx \amp = \int \frac{2x+4 - 14}{x^2+4x+5}\;dx\\ \amp = \int \frac{2x+4}{x^2+4x+5}\;dx - \int \frac{14}{x^2+4x+5}\;dx\\ \amp = \int \frac{1}{u};\du - \int \frac{14}{(x+2)^2+1}\;dx\\ \amp = \ln|x^2+4x+5| - 14\tan^{-1}(x+2) + C \end{align*}
So we end up with
\begin{equation*} 2\ln|x-1| - \int \frac{2x-10}{x^2+4x+5}\;dx = 2\ln|x-1| - \ln|x^2+4x+5| + 14\tan^{-1}(x+2) + C \end{equation*}

(f)

\(\dint \dfrac{x^2}{(x-2)^3}\;dx\)
Solution.
\begin{align*} \int \dfrac{x^2}{(x-2)^3}\;dx \amp = \int \frac{A}{(x-2)^3} + \frac{B}{(x-2)^2} + \frac{C}{x-2}\;dx\\ \amp = \int \frac{4}{(x-2)^3} + \frac{4}{(x-2)^2} + \frac{1}{x-2}\;dx\\ \amp = \int 4(x-2)^{-3} + 4(x-2)^{-2} + \frac{4}{x-2}\;dx\\ \amp = -2(x-2)^{-2} - 4(x-2)^{-1} + 4\ln|x-2| + C\\ \amp = 4\ln|x-2| - \frac{2}{(x-2)^2} - \frac{4}{x-2} + C \end{align*}

6.

In the problems we are looking at in this section, we’re limiting ourselves to, at most, irreducible quadratic factors in the denominator. In problems with simple linear factors, repeated linear factors, or irreducible quadratic factors, what types of antiderivative functions do you expect to see? Explain.

7.

For each of the following integrals, we will do some preliminary work before using partial fractions to integrate. Really, we’ll perform a specific \(u\)-substitution that will give us some resulting integral to use partial fractions on.

(a)

\(\dint \dfrac{4e^{2x}}{(e^{2x}+3)(e^{2x}-5)}\;dx\) where we use \(u=e^{2x}\text{.}\)
Hint.
If \(u=e^{2x}\) then \(du=2e^{2x}\;dx\) and our resulting integral looks like:
\begin{equation*} \int \frac{2}{(u+3)(u-5)}\;du\text{.} \end{equation*}
Solution.
Using \(u=e^{2x}\text{,}\) we get:
\begin{align*} \int \dfrac{4e^{2x}}{(e^{2x}+3)(e^{2x}-5)}\;dx \amp = \int \frac{2}{(u+3)(u-5)}\;du\\ \amp = \int \frac{A}{u+3} + \frac{B}{u-5} \;du\\ \amp = \int -\frac{1/4}{u+3} + \frac{1/4}{u-5}\;du\\ \amp = \frac{1}{4}\ln|u-5| - \frac{1}{4}\ln|u+3| + C\\ \amp = \frac{1}{4}\ln|e^{2x} - 5| - \frac{1}{4}\ln|e^{2x}+3| + C \end{align*}

(b)

\(\dint \sqrt{e^x+1}\;dx\) where we use \(u=\sqrt{e^x+1}\text{.}\)
Hint.
If \(u=\sqrt{e^x+1}\) then \(du = \dfrac{e^x}{2\sqrt{e^x+1}}\;dx\) and the resulting integral is:
\begin{equation*} \int \frac{2u^2}{u^2-1}\;du = \int 2 + \frac{2}{u^2-1}\;du\text{.} \end{equation*}
Solution.
Using \(u=\sqrt{e^x+1}\text{,}\) we get:
\begin{align*} \int \sqrt{e^x+1}\;dx \amp = \int \frac{2u^2}{u^2-1}\;du \\ \amp = \int 2 + \frac{2}{u^2-]1}\;du\\ \amp = \int 2 + \frac{2}{(u-1)(u+1)} \; du\\ \amp = \int 2 + \frac{A}{u-1} + \frac{B}{u+1}\;du\\ \amp = \int 2 + \frac{1}{u-1} - \frac{1}{u+1}\;du\\ \amp = 2u + \ln|u-1| - \ln|u+1| + C\\ \amp = 2\sqrt{e^x+1} + \ln|\sqrt{e^x+1}-1| - \ln|\sqrt{e^x+1}+1| + C \end{align*}

(c)

\(\dint \dfrac{\sqrt{x}+3}{\sqrt{x}(x-1)}\;dx\) where we use \(u=\sqrt{x}\text{.}\)
Hint.
If \(u=\sqrt{x}\) then \(du=\dfrac{1}{2\sqrt{x}}\;dx\) and the resulting integral is:
\begin{equation*} \int \frac{2u+6}{u^2-1}\;du\text{.} \end{equation*}
Solution.
Using \(u=\sqrt{x}\text{,}\) we get:
\begin{align*} \int \dfrac{\sqrt{x}+3}{\sqrt{x}(x-1)}\;dx \amp = \int \frac{2u+6}{u^2-1}\;du\\ \amp = \int \frac{2u+6}{u^2-1}\;du\\ \amp = \int \frac{2u+6}{(u-1)(u+1)}\;du\\ \amp = \int \frac{A}{u-1} + \frac{B}{u+1} \;du\\ \amp = \int \frac{4}{u-1} - \frac{2}{u+1} \;du\\ \amp = 4\ln|u-1| - 2\ln|u+1| + C\\ \amp = 4\ln|\sqrt{x}-1| - 2\ln|\sqrt{x} + 1| + C \end{align*}
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