When should I use partial fractions instead of u-substitution?

Use u-substitution when the numerator is a constant multiple of the derivative of the denominator, since that integrates directly to a logarithm. Use partial fractions when the denominator factors into distinct linear factors and no simple substitution collapses the whole fraction. If in doubt, test substitution first because it is faster.

Do I always need to factor the denominator?

Yes. Partial fraction decomposition requires the denominator to be expressed as a product of factors, because each factor becomes the denominator of one of the simpler fractions. If it is not factored, factor it before setting up the constants.

What if the numerator has a higher or equal degree than the denominator?

Then the fraction is improper and partial fractions cannot be applied directly. Do polynomial long division first to get a polynomial plus a proper remainder fraction, then decompose only the proper remainder.

Does AP Calculus BC test repeated or quadratic factors?

No. The AP scope for topic 6.12 limits partial fractions to distinct, non-repeated linear factors. You will not be asked to decompose denominators with repeated factors or irreducible quadratics on the exam.

AP-CALCBC-6.12

U6.12 Integration Using Partial Fractions

Master AP Calculus BC topic 6.12: decompose proper rational functions with linear factors into partial fractions and integrate each piece into logarithms.

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What this lesson covers

Some integrals of rational functions look impossible until you split them apart. The method of partial fractions rewrites a single messy fraction as a sum of simple fractions you already know how to integrate. On the AP Calculus BC exam, this technique appears with distinct linear factors in the denominator, and each piece integrates cleanly into a natural logarithm.

In this lesson you will learn how to check that a rational function is ready for decomposition, how to set up and solve for the unknown constants, and how to finish the integration. You will also see the warning signs that tell you to do something else first — like long division or substitution — before partial fractions can help.

When Partial Fractions Applies

Partial fraction decomposition works on a rational function

\frac{P(x)}{Q(x)}

, but only under two conditions you must always verify first.

First, the fraction must be proper: the degree of the numerator must be strictly less than the degree of the denominator. If it is not proper, do polynomial long division first (that connects to U6.10), then apply partial fractions to the leftover remainder term.

Second, the denominator must factor. On the AP exam the denominator factors into distinct linear factors, such as

(x-1)(x+3)

x(2x-1)

. Once factored, each distinct linear factor

(ax+b)

contributes one term of the form

\frac{A}{ax+b}

Situation	What to do
Degree of top $\geq$ degree of bottom	Long division first
Denominator not factored	Factor it
Distinct linear factors	Standard decomposition
Single factor, or $u$ -sub works	Skip partial fractions

A common misconception is reaching for partial fractions when a simple substitution would do. If the numerator is a constant multiple of the derivative of the denominator, use

u

-substitution instead — it is faster and avoids algebra.

Setting Up the Decomposition

Suppose you want to integrate

\frac{5x-1}{(x-1)(x+2)}

. Because both factors are linear and distinct, write

\frac{5x-1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.

Multiply both sides by the full denominator

(x-1)(x+2)

to clear fractions:

5x-1 = A(x+2) + B(x-1).

Now solve for the constants. The fastest approach is the cover-up method: substitute the value of

x

that makes each factor zero.

Let

x=1

5(1)-1 = A(1+2)

, so

4 = 3A

and

A = \frac{4}{3}

.

Let

x=-2

5(-2)-1 = B(-2-1)

, so

-11 = -3B

and

B = \frac{11}{3}

.

Alternatively, expand and match coefficients of like powers of

x

to build a system of equations. Both methods give the same answer; the cover-up method is usually quicker for distinct linear factors. Always double-check by recombining the fractions mentally to confirm the numerator matches.

Integrating the Pieces

Once decomposed, each term integrates into a natural logarithm using the pattern

\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C.

Continuing the previous setup:

\int \frac{5x-1}{(x-1)(x+2)}\,dx = \int \frac{4/3}{x-1}\,dx + \int \frac{11/3}{x+2}\,dx.

This equals

\frac{4}{3}\ln|x-1| + \frac{11}{3}\ln|x+2| + C.

Keep the absolute value bars — they matter because the logarithm is only defined for positive arguments, and the AP readers expect them. When the linear factor has a coefficient other than 1, such as

\frac{A}{2x-1}

, remember the extra

\frac{1}{2}

factor from the chain rule:

\int \frac{A}{2x-1}\,dx = \frac{A}{2}\ln|2x-1| + C

.

A frequent error is forgetting the constant of integration on an indefinite integral, or leaving off absolute values. For a definite integral, plug in bounds after integrating and simplify using logarithm properties like

\ln a - \ln b = \ln\frac{a}{b}

How the Exam Tests This

On the multiple-choice section, partial fractions questions are often short: recognize the structure, decompose, integrate, and match the logarithmic answer. Distractors typically differ by sign errors on the constants, missing coefficients like the

\frac{1}{a}

, or dropped absolute values.

On free-response questions, partial fractions frequently appears inside a larger problem — perhaps a differential equation from Unit 7 modeling logistic-style growth, or an area or volume computation. You may need to do long division first if the fraction is improper, so always check degrees before decomposing.

Exam clue	Likely technique
Factored quadratic denominator, constant top	Partial fractions
Numerator matches denominator's derivative	$u$ -substitution
Top degree $\geq$ bottom degree	Long division, then decide
Irreducible quadratic bottom	Complete the square (U6.10)

Because the AP scope covers only linear factors, you will not be asked to handle repeated or irreducible quadratic factors in the denominator on this exam. Practice recognizing the setup quickly so you spend your time on the algebra, not on deciding which method to use.

Key terms

Rational function.: A function written as a ratio of two polynomials, $\frac{P(x)}{Q(x)}$ , where $Q(x)$ is not zero.
Proper rational function.: A rational function whose numerator has degree strictly less than the denominator's degree; required before partial fraction decomposition.
Partial fraction decomposition.: Rewriting a proper rational function as a sum of simpler fractions, each with a single factor of the denominator underneath.
Distinct linear factors.: Denominator factors of the form $(ax+b)$ that each appear exactly once; the only case tested on AP Calculus BC.
Cover-up method.: A shortcut for finding decomposition constants by substituting the root of each linear factor into the cleared equation.
Polynomial long division.: The process used to reduce an improper rational function into a polynomial plus a proper remainder fraction before decomposing.

Worked example

Evaluate

\int \frac{3x+11}{x^2+x-6}\,dx

First check the degree: the numerator is degree 1 and the denominator is degree 2, so the fraction is proper and no long division is needed.

Factor the denominator:

x^2+x-6 = (x+3)(x-2)

. Both factors are linear and distinct, so set up

\frac{3x+11}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}.

Multiply through by

(x+3)(x-2)

3x+11 = A(x-2) + B(x+3).

Use the cover-up method. Let

x=2

3(2)+11 = B(2+3)

, so

17 = 5B

and

B = \frac{17}{5}

.

Let

x=-3

3(-3)+11 = A(-3-2)

, so

2 = -5A

and

A = -\frac{2}{5}

.

Now integrate each term:

\int \frac{-2/5}{x+3}\,dx + \int \frac{17/5}{x-2}\,dx = -\frac{2}{5}\ln|x+3| + \frac{17}{5}\ln|x-2| + C.

That is the final answer. Notice the absolute values and the single constant of integration.

Practice questions

Which of the following is the correct partial fraction setup for

\frac{7}{x(x-4)}

$\frac{A}{x} + \frac{B}{x-4}$
$\frac{A}{x} + \frac{Bx+C}{x-4}$
$\frac{Ax+B}{x(x-4)}$
$\frac{A}{x^2} + \frac{B}{x-4}$

Answer: $\frac{A}{x} + \frac{B}{x-4}$

The denominator has two distinct linear factors,

x

and

x-4

. Each distinct linear factor contributes exactly one term with a constant numerator, giving

\frac{A}{x} + \frac{B}{x-4}

. Linear-over-linear numerators like

Bx+C

are only used for irreducible quadratic factors, which are outside this scope.

Evaluate

\int \frac{1}{x^2-9}\,dx

Answer: $\frac{1}{6}\ln|x-3| - \frac{1}{6}\ln|x+3| + C$

Factor:

x^2-9 = (x-3)(x+3)

. Write

\frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}

, so

1 = A(x+3)+B(x-3)

. Let

x=3

1=6A

, giving

A=\frac{1}{6}

. Let

x=-3

1=-6B

, giving

B=-\frac{1}{6}

. Integrating each term yields

\frac{1}{6}\ln|x-3| - \frac{1}{6}\ln|x+3| + C

, which can also be written as

\frac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C

Before using partial fractions on

\int \frac{x^2+1}{x^2-x}\,dx

, what must you do first, and why?

Answer: Perform polynomial long division because the fraction is improper.

The numerator and denominator both have degree 2, so the fraction is not proper. Dividing gives

1 + \frac{x+1}{x^2-x}

. Now the remainder

\frac{x+1}{x(x-1)}

is proper and ready for decomposition. Skipping this step and decomposing directly produces an incorrect setup, because partial fractions requires the numerator degree to be strictly less than the denominator degree.

FAQ

When should I use partial fractions instead of u-substitution?: Use $u$ -substitution when the numerator is a constant multiple of the derivative of the denominator, since that integrates directly to a logarithm. Use partial fractions when the denominator factors into distinct linear factors and no simple substitution collapses the whole fraction. If in doubt, test substitution first because it is faster.
Do I always need to factor the denominator?: Yes. Partial fraction decomposition requires the denominator to be expressed as a product of factors, because each factor becomes the denominator of one of the simpler fractions. If it is not factored, factor it before setting up the constants.
What if the numerator has a higher or equal degree than the denominator?: Then the fraction is improper and partial fractions cannot be applied directly. Do polynomial long division first to get a polynomial plus a proper remainder fraction, then decompose only the proper remainder.
Does AP Calculus BC test repeated or quadratic factors?: No. The AP scope for topic 6.12 limits partial fractions to distinct, non-repeated linear factors. You will not be asked to decompose denominators with repeated factors or irreducible quadratics on the exam.

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