Hi everyone! Our integration technique toolbox is almost full. Today we’ll add the our last tool. We’ll see how this tool works in this lesson and in Lesson 9: Partial Fraction Decomposition (part 2) (link here).

Lesson 8: Partial Fraction Decomposition (part 1)

Learning goals:

  • Integrate a rational function using the method of partial fractions.
  • Recognize simple linear factors in a rational function.
  • Recognize repeated linear factors in a rational function.



  • Integration – Partial Fractions (complete some questions now, complete the rest after Lesson 9)

Motivating question and example

Question: What other algebraic strategies can we use to make evaluating integrals easier?

Example: How can we evaluate

\[\int \frac{3x+9}{x^2-4x-5}dx?\]

Warmup exercise 1



Show solution 1

By this point we’ve seen a number of integration techniques that we should consider to evaluate this integral. The integrand is a rational function and the denominator is quadratic, so you may be tempted to try out your trig substitution skills. But you should check first if there’s a simpler technique you can use! (You may like to re-watch Video 1 from Lesson 6 (link here) for a refresher; skip to minute 2:58.)

We’re lucky because we can integrate by direct substitution. We’ll set $u=x^2 – 4x – 5$ so $du = (2x-4)dx$:

\[\int\frac{2x-4}{x^2-4x-5}dx = \int \frac{du}{u}\]

\[ = \ln(|u|)+C \]

\[= \ln(|x^2 – 4x – 5|)+C.\]

Warmup exercise 2


\[\int \left (\frac{4}{x-5} – \frac{1}{x+1} \right)dx\]

Show solution 2

We can integrate term-by-term by direct substitution where $u$ is equal to $x-5$ and $x+1$ respectively, but since in both of these cases, $du = dx$, we can perform this substitution in our heads and integrate directly:

\[\int \left (\frac{4}{x-5} – \frac{1}{x+1} \right)dx = 4 \ln(|x-5|)- \ln(|x+1|+C.\]

Warmup exercise 3


\[\frac{4}{x-5} – \frac{1}{x+1}\]

Show solution 3

Our main task to subtract these fractions is to use the lowest common denominator. In this case, the LCD is $(x-5)(x+1)$. So:

\[\frac{4}{x-5} – \frac{1}{x+1} = \frac{4(x+1)}{(x-5)(x+1)} – \frac{x-5}{(x-5)(x+1)}\]



The question didn’t ask us to simplify the denominator, but we’ll use the simplified version later in the lesson, so we’ll simplify now:

\[\frac{4}{x-5} – \frac{1}{x+1} = \frac{3x+9}{x^2-4x-5}.\]

Partial fractions is an algebra technique, not a calculus technique

Here are three facts to notice about the warmup exercises:

  1. For warmup exercise 1, we used integration by direct substitution where $u$ was the denominator of the integrand $u= x^2-4x-5$ and $du= (2x-4)dx$, which is exactly the numerator (though if the numerator had been a constant multiple of $du$, we still could have used the same $u$) . This doesn’t help us with the motivating example, though, since the numerator $3x+9$ is not a constant multiple of $du$. The point of Warmup exercise 1 is to emphasize that we’re able to use such a direct substitution for rational integrands only in special cases. What should we do for other rational integrands? That’s the question that this lesson and the next one will answer.
  2. Warmup exercise 2 is pretty quick as long as we remember our integrals.
  3. Warmup exercise 3 is the key to unlocking the motivating example. Using Warmup exercise 3, we know that the integrand from the motivating example can be rewritten as the integrand from Warmup exercise 2. That means by completing Warmup exercise 2, we’ve actually already completed the motivating example! Notice that Warmup exercise 3 did not involve any calculus, just algebra!

Motivating example

\[\int \frac{3x+9}{x^2-4x-5}dx \stackrel{(\rm{how tho?})}{=} \int \left(\frac{4}{x-5} – \frac{1}{x+1}\right)dx\] \[= 4\ln(|x-5|) – \ln(|x+1|) + C.\]

The big question

So the integral from the motivating example became easy for us to evaluate when we had a simpler expression for the integrand. We saw that the two integrands in the motivating example and Warmup exercise 2 were equivalent when we combined the two terms from Warmup exercise 3 by finding the common denominator. But if we hadn’t been so lucky, what would we have done? How can we separate a rational function into a sum or difference of simpler ones? That is, how can we do something like Warmup exercise 3…in reverse?

Think of the partial fraction decomposition as the algebra technique that is the opposite of adding or subtracting rational functions by finding the common denominator. We’ve applied this algebra technique in the appropriate way if we’re left with something that’s easy to integrate.

Video 1 below walks us through the main details of the technique for today’s lesson and walks us through a few examples.

Video 1

More examples

Video 2 shows the application of this technique to the motivating example. Notice that we end up evaluating the integral from Warmup exercise 2, but this walks us through the process of separating the rational function into a sum or difference of simpler ones.

Video 2

Fundamental Theorem of Algebra

There are different versions of this theorem that you may have seen in your algebra or precalculus class. The version we’ll use is:

Theorem: Any polynomial with real coefficients can be factored uniquely into linear and irreducible quadratic factors.

Since we need to factor the denominator completely as the first step in finding the partial fraction decomposition of a rational function, the Fundamental Theorem of Algebra tells us we’ll can end up only with linear or irreducible quadratic factors. Today’s lesson will focus on denominators with linear factors only. We’ll see what to do with quadratic factors in Lesson 9 (link here).

We’ve already seen in Videos 1 and 2 what to do when the linear factor in question is unique. If the denominator has some power of a linear factor, we’ll have more terms in the partial fraction decomposition. Video 3 below shows how to handle these.

Video 3

Exit Ticket


\[\int \frac{5t+6}{t^2-36}dt\]