# Integration using partial fractions (BC only)

> AP Calculus BC · AP Calculus BC CED Unit 6: Integration and Accumulation of Change
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-using-partial-fractions/

This sub-topic covers integrating rational functions using partial fraction decomposition, an algebraic technique exclusive to AP Calculus BC. You will learn to handle proper/improper rationals, distinct and repeated linear factors, and avoid common exam errors.

**Prerequisites:** Integration of reciprocal linear and polynomial functions; Polynomial long division; Basic factoring of polynomials over the reals

## Learning objectives

- Distinguish between proper and improper rational functions
- Decompose rational functions with distinct and repeated linear factors
- Integrate decomposed rational functions using basic integration rules
- Identify and avoid common exam errors on this topic

## Proper vs Improper Rational Functions and Long Division

Partial fraction decomposition can only be applied directly to proper rational functions. If the function is improper ($\deg(P) \geq \deg(Q)$), 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 $\frac{2x^3 + x^2 - 5x + 2}{x^2 - 2x + 1}$ as the sum of a polynomial and a proper rational function.

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

> **tip**
>
> Always check the degrees of numerator and denominator before starting decomposition. Skipping long division for improper functions will lead to an incorrect decomposition and zero points on FRQ.

*Calculator:* allowed

## Partial Fractions with Distinct Linear Factors

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:

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

1. Check degrees: $\deg(\text{numerator}) = 1 < 2$, so the function is proper. Factor the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$, 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 $x=-1$: $5(-1) + 7 = A(1) \implies A=2$. Substitute $x=-2$: $5(-2) + 7 = B(-1) \implies B=3$.
5. Integrate term-by-term:
6. $$\int \left(\frac{2}{x+1} + \frac{3}{x+2}\right) dx = 2\ln|x+1| + 3\ln|x+2| + C = \ln\left|(x+1)^2(x+2)^3\right| + C$$

> **tip**
>
> Never omit absolute value bars inside the natural logarithm after integrating $\frac{1}{ax+b}$. AP exam graders require absolute value for full credit on FRQ.

*Calculator:* allowed

## Partial Fractions with Repeated Linear Factors

If a linear factor is repeated $k$ times in the denominator (i.e., $(ax+b)^k$ 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: $\int (ax+b)^{-k} dx = \frac{(ax+b)^{1-k}}{a(1-k)} + C$.

**Worked example:** Evaluate $\int \frac{3x^2 + 2x - 2}{x(x-1)^2} dx$.

1. Check degrees: $\deg(\text{numerator}) = 2 < 3$, 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 $x=0$: $-2 = A(1) \implies A=-2$. Substitute $x=1$: $3(1)^2 + 2(1) - 2 = C(1) \implies C=3$. Equate coefficients of $x^2$: $A + B = 3 \implies B=5$.
6. Integrate term-by-term:
7. $$\int \left(\frac{-2}{x} + \frac{5}{x-1} + \frac{3}{(x-1)^2}\right) dx = -2\ln|x| + 5\ln|x-1| - \frac{3}{x-1} + C$$

> **tip**
>
> Always add a term for every power of a repeated linear factor, including power 1. Skipping the power 1 term will result in an incorrect decomposition.

*Calculator:* allowed

## AP-Style Worked Practice Problems

**Worked example:** Evaluate $\int_3^4 \frac{2x}{x^2 - 3x + 2} dx$. Which option is equivalent to the result?

1. The integrand is proper, denominator factors to $(x-1)(x-2)$, distinct linear factors. Clear denominators: $2x = A(x-2) + B(x-1)$.
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 $f(x) = \frac{x^2 + 2x + 2}{x(x+1)^2}$ for $x>0$. (a) Find the partial fraction decomposition, (b) Find $\int f(x) dx$, (c) Find $g(2)$ if $g'(x)=f(x)$ and $g(1)=2\ln 2$.

1. (a) Set up decomposition, clear denominators: $x^2 + 2x + 2 = A(x+1)^2 + Bx(x+1) + Cx$. 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 $x>0$): $\int f(x) dx = 2\ln x - \ln(x+1) + \frac{1}{x+1} + C$.
3. (c) Use $g(1) = 2\ln 2$ to solve for $C = 3\ln 2 - \frac{1}{2}$. Substitute $x=2$: $g(2) = \ln\left(\frac{32}{3}\right) - \frac{1}{6}$.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Skipping long division when $\deg(P) \geq \deg(Q)$, e.g., trying to decompose $\frac{x^2}{x-1}$ directly as $\frac{A}{x-1}$.
  - Why it fails: Students rush to start partial fractions without checking degrees, forgetting decomposition only works for proper rationals.
  - Correct: Always write down the degree of numerator and denominator before starting; if $\deg(P) \geq \deg(Q)$, do long division first.
- **Wrong:** 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 it fails: Students confuse distinct and repeated factor rules, forgetting each power from 1 up to the exponent needs its own term.
  - Correct: For a repeated factor $(ax+b)^n$, write $n$ terms: $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_n}{(ax+b)^n}$.
- **Wrong:** 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 it fails: 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: Always add absolute value inside the logarithm when integrating any reciprocal linear function on the exam.
- **Wrong:** Sign errors when clearing denominators and solving for constants, e.g., getting $A=-2$ instead of $A=2$ after substituting a negative root.
  - Why it fails: Students rush substitution and do not check their work.
  - Correct: 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:** Factoring the denominator incorrectly, e.g., factoring $x^2 - 2x - 8$ as $(x-4)(x+1)$ instead of $(x-4)(x+2)$.
  - Why it fails: Students rush the factoring step, which undermines the entire problem.
  - Correct: After factoring, multiply the factors back to confirm you get the original denominator before setting up decomposition.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Proper Rational Function | $\frac{P(x)}{Q(x)}, \deg(P) < \deg(Q)$ | 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; $\deg(R) < \deg(Q)$ |
| Distinct Linear Factor Rule | $\frac{P(x)}{\prod (a_ix+b_i)} = \sum \frac{A_i}{a_ix+b_i}$ | Applies to all non-repeating linear factors |
| Repeated Linear Factor Rule | For $(ax+b)^k$, add $\sum_{i=1}^k \frac{A_i}{(ax+b)^i}$ | Add one term for every power from 1 to $k$ |
| Integral of $\frac{A}{ax+b}$ | $\int \frac{A}{ax+b} dx = A\ln\|ax+b\| + C$ | Requires absolute value inside the logarithm |
| Integral of $\frac{A}{(ax+b)^k}, k>1$ | $\int \frac{A}{(ax+b)^k} dx = \frac{A}{a(1-k)(ax+b)^{k-1}} + C$ | Use the power rule for this term |
| Root Substitution for Constants | Substitute $x = -b/a$ for $(ax+b)$ | Fastest method to solve for unknown constants |

## 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.

- [Improper integrals](https://www.owlsprep.com/study/ap-calculus-bc-u6-improper-integrals/)
- [Integration by parts](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-parts/)
- [Selecting techniques for antidifferentiation](https://www.owlsprep.com/study/ap-calculus-bc-u6-selecting-techniques-for-antidifferentiation/)

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