Linear partial fractions let you rewrite a complicated rational function as a sum of simpler fractions like $A/(x-r)$ , which are easy to integrate into natural log terms. You use this BC technique when the integrand is a proper rational function whose denominator factors into distinct, nonrepeating linear factors. For AP Calculus BC, factor the denominator first so each linear factor gets its own fraction.

Why This Matters for the AP Calculus Exam

This is a BC-only technique, so it shows up only on the AP Calculus BC exam, not AB. It supports the broader skill of selecting the right antidifferentiation method when you see a rational function you cannot integrate directly. On the exam you will often just be handed an integral and expected to recognize that decomposition is the move, then carry it out cleanly. Being fluent here also helps with later BC topics like improper integrals, where you may need to decompose first and then take a limit.

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Key Takeaways

Use partial fractions when the integrand is a rational function with a denominator that factors into distinct linear factors.
The numerator's degree must be less than the denominator's degree. If it is not, do polynomial long division first to make it proper.
Each distinct linear factor (ax + b) gets its own term of the form A/(ax + b).
Clear denominators, then solve for the constants by plugging in x-values that cancel the other terms (or by matching coefficients).
Each A/(x - r) term integrates to A ln|x - r|, so your answer is usually a sum of logarithms.
Include + C on indefinite integrals, and change limits properly if you switch to a substitution.

How to Use This on the AP Calculus Exam

Problem Solving

Work through partial fraction problems with a consistent routine:

Check that the integrand is a rational function with linear, nonrepeating factors in the denominator. That signals partial fraction decomposition.
If the numerator's degree is not less than the denominator's, do polynomial long division first.
Factor the denominator into distinct linear factors and split the fraction into a sum like A/(x - r) + B/(x - s).
Multiply both sides by the common denominator to clear fractions.
Solve for A, B, etc. by choosing x-values that cancel the other terms.
Plug the constants back in and integrate each simple fraction separately.

Worked Example: Already Factored

Start with this function:

R(x) = \cfrac{2x+1}{(x-1)(x+2)}

Split it into two simpler fractions with unknown numerators:

R(x)= \frac{A}{(x-1)} + \frac{B}{(x+2)}

Multiply both sides by the common denominator $(x-1)(x+2)$ to clear fractions:

2x+1 = A(x+2) + B(x-1)

Solve for A by substituting $x = 1$ , which cancels B:

2(1)+1=A(1+2)+B(1-1)

2+1=3A

A=1

Solve for B by substituting $x = -2$ , which cancels A:

2(-2)+1=A(-2+2)+B(-2-1)

-4+1=-3B

B=1

Put the constants back in:

R(x)= \frac{1}{(x-1)}+\frac{1}{(x+2)}

Now integrate each term. If you want a refresher on the basic log antiderivatives, see the topic 6.8 guide. The result is:

\intop R(x)\,dx=\ln|x-1|+\ln|x+2|+C

Worked Example: Factor First

Real problems often hand you an unfactored denominator, so sharp factoring saves time. Consider:

R(x)=\frac{3x+2}{x^2-13x+42}

Factor the denominator:

R(x)=\frac{3x+2}{(x-6)(x-7)}

Set up the decomposition and clear denominators:

R(x)=\frac{A}{x-6}+\frac{B}{x-7}

3x+2=A(x-7)+B(x-6)

Solve for A by substituting $x = 6$ :

3(6)+2=A(6-7)+B(6-6)

A=-20

Solve for B by substituting $x = 7$ :

3(7)+2=A(7-7)+B(7-6)

B=23

Integrate each term:

R(x)=\frac{-20}{x-6}+\frac{23}{x-7}

\intop R(x)\,dx=-20\ln|x-6| + 23 \ln|x-7|+C

Common Trap

When you are given an integral with no instructions, the cue for partial fractions is a rational function where the numerator's degree is lower than the denominator's and the denominator factors into distinct linear pieces. If the numerator's degree is too high, the integrand is improper, so divide first. If the factoring does not give clean linear factors, a different technique may fit better.

Common Misconceptions

Forgetting to check the degree first. Partial fractions only apply directly to proper rational functions. If the numerator's degree is greater than or equal to the denominator's, you must do polynomial long division before decomposing.
Dropping the + C. Indefinite integrals need a constant of integration. Leaving it off costs you on free-response work.
Thinking every rational function decomposes into simple logs. This topic focuses on distinct, nonrepeating linear factors. Other denominators behave differently, so confirm the factors are linear and distinct before using this method.
Sign errors when solving for constants. When you substitute x-values to cancel terms, watch the signs carefully, especially with negative roots like $x = -2$ .
Forgetting to switch back after integrating. Each A/(x - r) becomes A ln|x - r|. Keep the absolute value bars so the antiderivative is valid where the function is defined.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.
indefinite integral	Antiderivatives of a function, represented as ∫f(x)dx = F(x) + C, where C is an arbitrary constant.
linear factors	First-degree polynomial expressions of the form (ax + b) used as denominators in partial fraction decomposition.
linear partial fractions	A decomposition technique that expresses a rational function as a sum of simpler fractions with linear denominators, used to simplify integration.
nonrepeating factors	Linear factors that appear only once in the denominator of a rational function, used in partial fraction decomposition.
rational function	A function expressed as the ratio of two polynomial functions.

Frequently Asked Questions

What are linear partial fractions?

Linear partial fractions rewrite a rational function as a sum of simpler fractions whose denominators are linear factors. This makes the function easier to integrate.

When should I use partial fractions in AP Calculus BC?

Use partial fractions when the integrand is a rational function and the denominator factors into linear pieces. If the fraction is improper, do polynomial long division first.

How do you set up a partial fraction decomposition?

Factor the denominator, assign a constant numerator over each distinct linear factor, clear denominators, then solve for the constants.

How do partial fractions integrate?

After decomposition, each simple fraction usually integrates to a natural log expression, such as a constant times ln absolute value of a linear factor.

Is linear partial fractions tested on AP Calculus AB?

No. Linear partial fractions is an AP Calculus BC topic, not an AP Calculus AB topic.

What is the biggest mistake with partial fractions?

The most common mistake is decomposing before checking that the rational function is proper. If the numerator degree is too high, divide first.