6.12 Integrating Using Linear Partial Fractions

BC only. When a fraction's denominator factors into different linear factors, you can split it into simpler fractions A/(x − a) + B/(x − b), each of which integrates to a natural log. This technique is called partial fraction decomposition.

Key terms

partial fraction decomposition
linear factors
rational function
natural logarithm

The idea

This topic is tested only on the BC exam.

Adding fractions goes one way: 2/(x − 2) + 3/(x + 1) = (5x − 4)/((x − 2)(x + 1)). Partial fractions goes the other way. You start with the combined fraction and recover the simple pieces, because each piece is easy to integrate: ∫ A/(x − a) dx = A ln|x − a| + C.

On the AP exam you only need distinct (non-repeating) linear factors, like (x − 2)(x + 1) or x(x + 3)(x − 1). Repeated factors such as (x − 1)² and irreducible quadratic factors such as (x² + 1) aren't tested.

The steps

Make sure the top's degree is less than the bottom's. If not, divide first (6.10).
Factor the denominator completely into different linear factors.
Write one unknown constant over each factor: P(x)/((x − a)(x − b)) = A/(x − a) + B/(x − b).
Multiply both sides by the full denominator: P(x) = A(x − b) + B(x − a).
Find the constants. The fastest way: plug in x = a (which wipes out the B term) and x = b (which wipes out the A term).
Integrate each piece as a natural log.

Two ways to find the constants

Plugging in roots (the method above) is quickest. You can also expand and match coefficients: for 5x − 4 = A(x + 1) + B(x − 2), expand to (A + B)x + (A − 2B), then solve A + B = 5 and A − 2B = −4. Both give A = 2 and B = 3.

Always check by recombining or by plugging in an easy x-value, like x = 0, into both forms.

With three linear factors, the setup is the same: P(x)/((x − a)(x − b)(x − c)) = A/(x − a) + B/(x − b) + C/(x − c), and plugging in a, b and c gives each constant in one step.

Where it shows up

Partial fractions appears in plain antiderivative questions, in definite and improper integrals (6.13), and behind the solution of the logistic differential equation (7.9). It's also how you'd show that a telescoping series like Σ 1/(n(n + 1)) collapses (10.1), because 1/(n(n + 1)) = 1/n − 1/(n + 1).

Worked examples

Try each one yourself first, then open the solution.

Example 1
Basic decomposition
Find ∫ (5x − 4)/((x − 2)(x + 1)) dx.
Show the solutionHide the solution
1. Step 1: Write (5x − 4)/((x − 2)(x + 1)) = A/(x − 2) + B/(x + 1).
2. Step 2: Multiply through: 5x − 4 = A(x + 1) + B(x − 2).
3. Step 3: x = 2: 6 = 3A, so A = 2.
4. Step 4: x = −1: −9 = −3B, so B = 3.
5. Step 5: Integrate: ∫ [2/(x − 2) + 3/(x + 1)] dx = 2 ln|x − 2| + 3 ln|x + 1| + C.
Answer: 2 ln|x − 2| + 3 ln|x + 1| + C
Example 2
Definite integral
Evaluate ∫₂⁴ 1/(x² − 1) dx.
Show the solutionHide the solution
1. Step 1: Factor: x² − 1 = (x − 1)(x + 1). Write 1 = A(x + 1) + B(x − 1).
2. Step 2: x = 1: 1 = 2A, A = ½. x = −1: 1 = −2B, B = −½.
3. Step 3: Antiderivative: ½ ln|x − 1| − ½ ln|x + 1| = ½ ln|(x − 1)/(x + 1)|.
4. Step 4: At 4: ½ ln(3/5). At 2: ½ ln(1/3).
5. Step 5: Subtract: ½[ln(3/5) − ln(1/3)] = ½ ln(9/5).
Answer: ½ ln(9/5) ≈ 0.294
Example 3
Trap: divide first
Find ∫ (x² + 1)/(x² − x) dx.
Show the solutionHide the solution
1. Step 1: The degrees are equal (2 and 2), so partial fractions can't start yet. Divide: (x² + 1)/(x² − x) = 1 + (x + 1)/(x² − x).
2. Step 2: Factor: x² − x = x(x − 1). Write x + 1 = A(x − 1) + Bx.
3. Step 3: x = 0: 1 = −A, A = −1. x = 1: 2 = B.
4. Step 4: Integrand: 1 − 1/x + 2/(x − 1). Integrate: x − ln|x| + 2 ln|x − 1| + C.
Answer: x − ln|x| + 2 ln|x − 1| + C

Common mistakes

Skipping the long division when the top's degree is at least the bottom's.
Plugging a root into the wrong factor and getting a sign wrong. Write the equation P(x) = A(…) + B(…) out fully before plugging in.
Using partial fractions when u-substitution is quicker: ∫ 2x/(x² − 1) dx = ln|x² − 1| + C directly.
Dropping the absolute values in the logs.

On the exam

BC multiple-choice questions often give an integral with a factorable quadratic denominator and answer choices made of logs. Find A and B, then match.
Partial fractions sometimes appears in an improper integral, where you take a limit of a log expression. Combine the logs into one log of a fraction before taking the limit.

Connected topics

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