Integration using partial fractions (BC only) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Decomposition of proper and improper rational functions, partial fraction decomposition for distinct linear factors, repeated linear factors, and integration of all resulting forms per AP Calculus BC CED requirements.

You should already know: Integration of reciprocal linear and polynomial functions; polynomial long division; basic factoring of polynomials over the reals.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Integration using partial fractions (BC only)?

Integration by partial fractions is an algebraic integration technique exclusively for rational functions, which are ratios of two polynomials written as $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ have real coefficients. It works by decomposing a complex rational integrand into a sum of simpler rational functions that have known antiderivatives, so you can integrate term-by-term using basic rules. This topic is only tested on AP Calculus BC, not AB, and per the College Board CED, it accounts for approximately 2-4% of the total BC exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it may be a standalone MCQ integration problem, or a required intermediate step in larger FRQ problems involving accumulated change, differential equations, area, or volume. The method is fully deterministic for any rational function with a factorable denominator, relying on basic algebra rather than trial and error, so it is very learnable with practice.

2. Proper vs Improper Rational Functions and Long Division

Partial fraction decomposition can only be applied directly to proper rational functions, which are rational functions where the degree of the numerator $\mathrm{deg}(P)$ is strictly less than the degree of the denominator $\mathrm{deg}(Q)$. If the function is improper ($\mathrm{deg}(P)\ge \mathrm{deg}(Q)$), you must first rewrite it as the sum of a polynomial and a new proper rational function using polynomial long division. The polynomial term can be integrated directly using the power rule, and only the proper remainder term requires partial fraction decomposition. For AP Calculus BC, you will only ever encounter linear factors (distinct or repeated) in the denominator on the exam; irreducible quadratic factors are not tested per the CED, so you do not need to practice that case. After long division, the next step is to factor the denominator of the proper remainder term completely into linear factors before starting decomposition.

Worked Example

Problem: 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: $\mathrm{deg}(\text{numerator})=3$, $\mathrm{deg}(\text{denominator})=2$, so $3\ge 2$: the function is improper, so 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 $2x$ to get $2x^3-4x^2+2x$, and subtract from the dividend to get $5x^2-7x+2$.
3. The next term of the quotient is $+5$; multiply the divisor by $5$ to get $5x^2-10x+5$, and subtract to get the remainder $3x-3$, which has degree 1 < 2, so we stop.
4. Rewrite the original function: $$\frac{2x^3+x^2-5x+2}{x^2-2x+1}=2x+5+\frac{3x-3}{(x-1{)}^2}=2x+5+\frac{3}{x-1}$$

Exam 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.

3. Partial Fractions with Distinct Linear Factors

After obtaining a proper rational function, the most common case you will encounter on the exam is a denominator that factors into distinct (non-repeating) linear factors. That means the denominator can be written as $Q(x)=(a_{1}x+b_1)(a_{2}x+b_2)...(a_{n}x+b_n)$ where no factor is repeated. For this case, the partial fraction decomposition follows the rule: $$\frac{P(x)}{Q(x)}=\frac{A_1}{a_{1}x+b_1}+\frac{A_2}{a_{2}x+b_2}+...+\frac{A_n}{a_{n}x+b_n}$$ where $A_1,A_2,...A_n$ are unknown constant coefficients. To solve for the constants, multiply both sides by $Q(x)$ to eliminate all denominators. The fastest method for distinct linear factors is root substitution: substitute the root of each linear factor (i.e., $x=-b_i/a_i$) into the resulting equation to solve for $A_i$ directly, since all other terms become zero. Once you have all constants, integrate each term individually using the rule for integrating reciprocal linear functions.

Worked Example

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

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

Exam 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.

4. Partial Fractions with Repeated Linear Factors

If a linear factor is repeated $k$ times in the denominator (i.e., it appears as $(ax+b{)}^k$ for $k\ge 2$), you cannot just write a single term $\frac{A}{(ax+b{)}^k}$ in your decomposition. Instead, 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 $\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}$. To solve for constants, you can still use root substitution to find the constant for the highest power of the repeated factor, then use equating coefficients of like powers of $x$ to find the remaining constants. When integrating, terms with power 1 still integrate to a logarithm, while terms with power $k\ge 2$ integrate using the power rule, since $\int (ax+b{)}^{-k}dx=\frac{(ax+b{)}^{1-k}}{a(1-k)}+C$.

Worked Example

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

1. Check degrees: $\mathrm{deg}(\text{numerator})=2<3$, so proper. Denominator has one distinct linear factor $x$ and one repeated linear factor $(x-1{)}^2$.
2. Set up decomposition and clear denominators: $$\frac{3x^2+2x-2}{x(x-1{)}^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}$$ $$3x^2+2x-2=A(x-1{)}^2+Bx(x-1)+Cx$$
3. 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 $x^2$: left side coefficient is 3, right side coefficient is $A+B=3$, so $-2+B=3⟹B=5$.
4. Integrate term-by-term: $$\int \left(\frac{-2}{x}+\frac{5}{x-1}+\frac{3}{(x-1{)}^2}\right)dx=-2\ln |x|+5\ln |x-1|-\frac{3}{x-1}+C$$

Exam tip: Always add a term for every power of a repeated linear factor, including power 1. Skipping the power 1 term (e.g., only writing $\frac{C}{(x-1{)}^2}$) will result in an incorrect decomposition.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Skipping long division when the numerator degree is greater than or equal to the denominator degree, 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 $\mathrm{deg}(P)\ge \mathrm{deg}(Q)$, 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, and forget each power from 1 up to the exponent needs its own term. Correct move: 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 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 $x^2-2x-8$ 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Evaluate ${\int }_3^4\frac{2x}{x^2-3x+2}dx$. Which of the following is equivalent to the result? A) $\ln \left(\frac{3}{2}\right)$ B) $4\ln 2-2\ln 3$ C) $2\ln 3-3\ln 2$ D) $\ln 4-1$

Worked Solution: The integrand is proper, with denominator factoring to $(x-1)(x-2)$, two distinct linear factors. We set up decomposition to get $2x=A(x-2)+B(x-1)$, then solve: substituting $x=1$ gives $A=-2$, substituting $x=2$ gives $B=4$. The antiderivative is $-2\ln |x-1|+4\ln |x-2|+C$. Evaluate from 3 to 4: $(-2\ln 3+4\ln 2)-(-2\ln 2+4\ln 1)=4\ln 2-2\ln 3$. This matches option B. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{x^2+2x+2}{x(x+1{)}^2}$ for $x>0$. (a) Find the partial fraction decomposition of $f(x)$. (b) Find $\int f(x)dx$, leaving your answer in simplest form with constant of integration $C$. (c) Given that $f(x)$ is the derivative of $g(x)$, and $g(1)=2\ln 2$, find $g(2)$.

Worked Solution: (a) $f(x)$ is proper. Set up decomposition: $$\frac{x^2+2x+2}{x(x+1{)}^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$$ Clear denominators: $x^2+2x+2=A(x+1{)}^2+Bx(x+1)+Cx$. Substitute $x=0$ to get $A=2$, substitute $x=-1$ to get $C=-1$, equate $x^2$ coefficients to get $2+B=1⟹B=-1$. Decomposition: $\boxed{\frac{2}{x}-\frac{1}{x+1}-\frac{1}{(x+1{)}^2}}$.

(b) Integrate term-by-term: $$\int f(x)dx=\int \left(\frac{2}{x}-\frac{1}{x+1}-\frac{1}{(x+1{)}^2}\right)dx=\boxed{2\ln x-\ln (x+1)+\frac{1}{x+1}+C}$$ (absolute values are omitted since $x>0$).

(c) Use $g(1)=2\ln 2$: $$g(1)=2\ln 1-\ln 2+\frac{1}{2}+C=-\ln 2+\frac{1}{2}+C=2\ln 2⟹C=3\ln 2-\frac{1}{2}$$ Substitute $x=2$: $$g(2)=2\ln 2-\ln 3+\frac{1}{3}+3\ln 2-\frac{1}{2}=\boxed{5\ln 2-\ln 3-\frac{1}{6}=\ln \left(\frac{32}{3}\right)-\frac{1}{6}}$$

Question 3 (Application / Real-World Style)

The rate of change of the number of bacteria in a culture after $t\ge 0$ hours is given by $\frac{dN}{dt}=\frac{100t}{(t+1)(t+2)}$ bacteria per hour. Find the total change in the number of bacteria between $t=0$ and $t=3$ hours, rounded to the nearest whole number.

Worked Solution: The total change is given by the definite integral ${\int }_0^3\frac{100t}{(t+1)(t+2)}dt$. Decompose the integrand: $\frac{t}{(t+1)(t+2)}=\frac{-1}{t+1}+\frac{2}{t+2}$. Integrate: $$100{\int }_0^3\left(\frac{-1}{t+1}+\frac{2}{t+2}\right)dt=100{\left[-\ln (t+1)+2\ln (t+2)\right]}_0^3$$ Evaluate the bounds: $$100\left[\left(-\ln 4+2\ln 5\right)-\left(-\ln 1+2\ln 2\right)\right]=100\left(\ln 25-4\ln 2\right)\approx 100(3.2189-2.7726)=44.63$$ Interpretation: The total number of bacteria in the culture increases by approximately 45 bacteria over the first 3 hours of the experiment.

7. Quick Reference Cheatsheet

8. 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 FRQ. 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 problems for full credit on the exam. After mastering partial fractions, you will move on to other advanced integration topics in Unit 6 and beyond. Review the following related topics next: Improper integrals Integration by parts Separable differential equations Accumulation of change