Integration using partial fractions (BC only)

AP Calculus BC· AP Calculus BC CED — Integration and Accumulation of Change· 14 min read

1. Proper vs Improper Rational Functions and Long Division★★☆☆☆BC only⏱ 3 min

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Partial fraction decomposition can only be applied directly to proper rational functions. If the function is improper ( $de g (P) \geq de g (Q)$ ), you must first rewrite it as the sum of a polynomial and a new proper rational function using polynomial long division. For AP Calculus BC, you will only encounter linear factors (distinct or repeated) in the denominator; irreducible quadratic factors are not tested.

📐 Worked Example

Rewrite $\frac{2 x ^{3} + x ^{2} - 5 x + 2}{x ^{2} - 2 x + 1}$ as the sum of a polynomial and a proper rational function.

1
Check degrees: $de g (numerator) = 3$ , $de g (denominator) = 2$ , so $3 \geq 2$ : the function is improper, long division is required.
2
Divide $2 x^{3} + x^{2} - 5 x + 2$ by $x^{2} - 2 x + 1$ . The first term of the quotient is $2 x$ ; multiply the divisor by $2 x$ and subtract to get $5 x^{2} - 7 x + 2$ .
3
The next term of the quotient is $+ 5$ ; multiply the divisor by $5$ and subtract to get the remainder $3 x - 3$ , which has degree $1 < 2$ , so we stop.
4
Rewrite the original function:
5
$\frac{2x^3 + x^2 - 5x + 2}{x^2 - 2x + 1} = 2x + 5 + \frac{3x - 3}{(x-1)^2} = 2x + 5 + \frac{3}{x-1}$

2. Partial Fractions with Distinct Linear Factors★★★☆☆BC only⏱ 4 min

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The most common case on the exam is a denominator that factors into distinct (non-repeating) linear factors. For this case, the decomposition follows the rule:

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To solve for constants, multiply both sides by $Q (x)$ to eliminate denominators. The fastest method is root substitution: substitute the root of each linear factor into the resulting equation to solve for $A_{i}$ directly, since all other terms become zero. Integrate term-by-term once you have all constants.

📐 Worked Example

Evaluate $\int \frac{5 x + 7}{x ^{2} + 3 x + 2} d x$ .

1
Check degrees: $de g (numerator) = 1 < 2$ , so the function is proper. Factor the denominator: $x^{2} + 3 x + 2 = (x + 1) (x + 2)$ , distinct linear factors.
2
Set up decomposition and clear denominators:
3
$\frac{5x+7}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \implies 5x + 7 = A(x+2) + B(x+1)$
4
Solve for constants with root substitution: Substitute $x = - 1$ : $5 (- 1) + 7 = A (1) ⟹ A = 2$ . Substitute $x = - 2$ : $5 (- 2) + 7 = B (- 1) ⟹ B = 3$ .
5
Integrate term-by-term:
6
$\int \left(\frac{2}{x+1} + \frac{3}{x+2}\right) dx = 2\ln|x+1| + 3\ln|x+2| + C = \ln\left|(x+1)^2(x+2)^3\right| + C$

3. Partial Fractions with Repeated Linear Factors★★★★☆BC only⏱ 4 min

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If a linear factor is repeated $k$ times in the denominator (i.e., $(a x + b)^{k}$ for $k \geq 2$ ), you must add a separate term for every power of the factor from 1 up to $k$ . For example, if the denominator is $x (x - 1)^{2}$ , the decomposition is $\frac{A}{x} + \frac{B}{x - 1} + \frac{C}{( x - 1 ) ^{2}}$ .

When integrating, terms with power 1 still integrate to a logarithm, while terms with power $k \geq 2$ integrate using the power rule: $\int (a x + b)^{- k} d x = \frac{( a x + b ) ^{1 - k}}{a ( 1 - k )} + C$ .

📐 Worked Example

Evaluate $\int \frac{3 x ^{2} + 2 x - 2}{x ( x - 1 ) ^{2}} d x$ .

1
Check degrees: $de g (numerator) = 2 < 3$ , so the function is proper. Denominator has one distinct factor $x$ and one repeated factor $(x - 1)^{2}$ .
2
Set up decomposition and clear denominators:
3
$\frac{3x^2 + 2x - 2}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$
4
$3x^2 + 2x - 2 = A(x-1)^2 + Bx(x-1) + Cx$
5
Solve for constants: Substitute $x = 0$ : $- 2 = A (1) ⟹ A = - 2$ . Substitute $x = 1$ : $3 (1)^{2} + 2 (1) - 2 = C (1) ⟹ C = 3$ . Equate coefficients of $x^{2}$ : $A + B = 3 ⟹ B = 5$ .
6
Integrate term-by-term:
7
$\int \left(\frac{-2}{x} + \frac{5}{x-1} + \frac{3}{(x-1)^2}\right) dx = -2\ln|x| + 5\ln|x-1| - \frac{3}{x-1} + C$

4. AP-Style Worked Practice Problems★★★★☆BC only⏱ 3 min

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📐 Worked Example

Evaluate $\int_{3}^{4} \frac{2 x}{x ^{2} - 3 x + 2} d x$ . Which option is equivalent to the result?

1
The integrand is proper, denominator factors to $(x - 1) (x - 2)$ , distinct linear factors. Clear denominators: $2 x = A (x - 2) + B (x - 1)$ .
2
Solve for constants: $x = 1$ gives $A = - 2$ , $x = 2$ gives $B = 4$ . Antiderivative is $- 2 ln ∣ x - 1∣ + 4 ln ∣ x - 2∣ + C$ .
3
Evaluate from 3 to 4: $(- 2 ln 3 + 4 ln 2) - (- 2 ln 2 + 4 ln 1) = 4 ln 2 - 2 ln 3$ . The correct answer is B.

📐 Worked Example

Let $f (x) = \frac{x ^{2} + 2 x + 2}{x ( x + 1 ) ^{2}}$ for $x > 0$ . (a) Find the partial fraction decomposition, (b) Find $\int f (x) d x$ , (c) Find $g (2)$ if $g^{'} (x) = f (x)$ and $g (1) = 2 ln 2$ .

1
(a) Set up decomposition, clear denominators: $x^{2} + 2 x + 2 = A (x + 1)^{2} + B x (x + 1) + C x$ . Substitute $x = 0$ to get $A = 2$ , $x = - 1$ to get $C = - 1$ , equate $x^{2}$ coefficients to get $B = - 1$ . Decomposition: $\frac{2}{x} - \frac{1}{x + 1} - \frac{1}{( x + 1 ) ^{2}}$ .
2
(b) Integrate term-by-term (absolute values omitted since $x > 0$ ): $\int f (x) d x = 2 ln x - ln (x + 1) + \frac{1}{x + 1} + C$ .
3
(c) Use $g (1) = 2 ln 2$ to solve for $C = 3 ln 2 - \frac{1}{2}$ . Substitute $x = 2$ : $g (2) = ln (\frac{32}{3}) - \frac{1}{6}$ .

5. Common Pitfalls

Wrong move:

Skipping long division when $de g (P) \geq de g (Q)$ , e.g., trying to decompose $\frac{x ^{2}}{x - 1}$ directly as $\frac{A}{x - 1}$ .

Why:

Students rush to start partial fractions without checking degrees, forgetting decomposition only works for proper rationals.

Correct move:

Always write down the degree of numerator and denominator before starting; if $de g (P) \geq de g (Q)$ , do long division first.

Wrong move:

Only writing one term for a repeated linear factor, e.g., decomposing $\frac{P ( x )}{x ( x - 2 ) ^{2}}$ as $\frac{A}{x} + \frac{B}{( x - 2 ) ^{2}}$ .

Why:

Students confuse distinct and repeated factor rules, forgetting each power from 1 up to the exponent needs its own term.

Correct move:

For a repeated factor $(a x + b)^{n}$ , write $n$ terms: $\frac{A _{1}}{a x + b} + \frac{A _{2}}{( a x + b ) ^{2}} + ... + \frac{A _{n}}{( a x + b ) ^{n}}$ .

Wrong move:

Dropping absolute value in the logarithm after integrating $\frac{1}{x + c}$ , e.g., writing $ln (x + 1) + C$ instead of $ln ∣ x + 1∣ + C$ .

Why:

Students remember the antiderivative of $\frac{1}{x}$ is $ln x$ from early lessons, forgetting $ln x$ is only defined for positive $x$ , but $\frac{1}{x}$ is defined for negative $x$ too.

Correct move:

Always add absolute value inside the logarithm when integrating any reciprocal linear function on the exam.

Wrong move:

Sign errors when clearing denominators and solving for constants, e.g., getting $A = - 2$ instead of $A = 2$ after substituting a negative root.

Why:

Students rush substitution and do not check their work.

Correct move:

After solving for all constants, plug them back into the cleared equation and test with any non-root $x$ value to confirm both sides match before integrating.

Wrong move:

Factoring the denominator incorrectly, e.g., factoring $x^{2} - 2 x - 8$ as $(x - 4) (x + 1)$ instead of $(x - 4) (x + 2)$ .

Why:

Students rush the factoring step, which undermines the entire problem.

Correct move:

After factoring, multiply the factors back to confirm you get the original denominator before setting up decomposition.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Proper Rational Function	$\frac{P ( x )}{Q ( x )}, de g (P) < de g (Q)$	Can be decomposed directly without long division
Improper Rational Function	$\frac{P ( x )}{Q ( x )} = S (x) + \frac{R ( x )}{Q ( x )}$	Use polynomial long division; $de g (R) < de g (Q)$
Distinct Linear Factor Rule	$\frac{P ( x )}{\prod ( a _{i} x + b _{i} )} = \sum \frac{A _{i}}{a _{i} x + b _{i}}$	Applies to all non-repeating linear factors
Repeated Linear Factor Rule	For $(a x + b)^{k}$ , add $\sum_{i = 1}^{k} \frac{A _{i}}{( a x + b ) ^{i}}$	Add one term for every power from 1 to $k$
Integral of $\frac{A}{a x + b}$	$\int \frac{A}{a x + b} d x = A ln ∣ a x + b ∣ + C$	Requires absolute value inside the logarithm
Integral of $\frac{A}{( a x + b ) ^{k}}, k > 1$	$\int \frac{A}{( a x + b ) ^{k}} d x = \frac{A}{a ( 1 - k ) ( a x + b ) ^{k - 1}} + C$	Use the power rule for this term
Root Substitution for Constants	Substitute $x = - b / a$ for $(a x + b)$	Fastest method to solve for unknown constants

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Standalone definite integral problem
2023 · FRQ
Intermediate step for differential equation

What's Next

Integration using partial fractions is a core prerequisite for solving separable differential equations with rational right-hand sides, which frequently appear on AP Calculus BC free-response questions. It is also a common intermediate step when computing volumes of revolution, finding accumulated change of a rational rate function, and evaluating definite integrals for contextual problems. Without mastering this decomposition technique, you will not be able to complete these multi-step problems for full credit on the exam. After mastering partial fractions, you can move on to other advanced integration topics and applications in Unit 6 and beyond.