Difficulty: Intermediate | Prerequisites: Factoring polynomials, basic integration (ln, power rule), polynomial long division
Partial fraction decomposition is an integration technique you reach for when you have a rational function (a polynomial divided by a polynomial) and none of the simpler methods work. It sits in the Calc 2 integration toolkit alongside substitution, integration by parts, and trigonometric integrals. The core idea is to break a complicated fraction into a sum of simpler fractions whose integrals you already know, typically producing natural-log terms. You should be comfortable factoring quadratics and polynomials before diving in, and you will use polynomial long division whenever the fraction is improper.
To integrate a rational function P(x)/Q(x), factor the denominator, split the fraction into simpler pieces with unknown constants, solve for those constants, then integrate each piece. If the degree of the top is not less than the degree of the bottom, do long division first to make it proper.
Rational function
A function of the form P(x)/Q(x) where both P and Q are polynomials. Think of it as one polynomial sitting on top of another, written as a fraction.
Proper rational function
A rational function where the degree of the numerator is strictly less than the degree of the denominator. In simple terms, the top polynomial is "smaller" than the bottom one.
Improper rational function
A rational function where the degree of the numerator is greater than or equal to the degree of the denominator. You must use long division to convert it before applying partial fractions.
Partial fraction decomposition
The process of rewriting a single complicated fraction as a sum of simpler fractions whose denominators are the factors of the original denominator. Think of it as "un-adding" fractions, the reverse of finding a common denominator.
Irreducible quadratic factor
A quadratic expression (ax^2 + bx + c) that cannot be factored into real linear factors because its discriminant is negative. In simple terms, it has complex roots and will not break down further over the reals.
Repeated factor
A factor that appears more than once in the denominator, such as (x - 1)^2 or (x^2 + 4)^2. Each power of the repeated factor gets its own term in the decomposition.
Check the degree. If deg(P) >= deg(Q), perform polynomial long division first to get a polynomial plus a proper fraction. Only the proper fraction needs decomposition.
Factor Q(x) completely. Break the denominator into linear factors (x - a) and irreducible quadratic factors (x^2 + bx + c). Note any repeated factors.
Set up the decomposition. Write one term for each factor, with unknown constants in the numerator. The form of each term depends on the factor type (see the four cases below).
Clear the denominator. Multiply both sides by Q(x) to get a polynomial equation.
Solve for the constants. Use either of two methods:
Strategic substitution: plug in the roots of Q(x) to zero out terms and isolate constants one at a time.
Equating coefficients: expand the right side, group by powers of x, and match coefficients with the left side to build a system of equations.
Integrate each simple fraction. Linear denominators give ln terms. Quadratic denominators give ln and/or arctan terms.
Integrate (3x + 4) / (x^2 + 3x + 2) dx.
Factor the bottom: x^2 + 3x + 2 = (x + 1)(x + 2)
Set up: (3x + 4) / [(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)
Multiply through: 3x + 4 = A(x + 2) + B(x + 1)
Expand: 3x + 4 = (A + B)x + (2A + B)
Match coefficients: A + B = 3 and 2A + B = 4, giving A = 1, B = 2
Integrate: integral = ln|x + 1| + 2 ln|x + 2| + C
The simplest case. Q(x) factors as a product of different linear terms, and none repeat.
Example denominator: Q(x) = x(x - 1)(x + 2)(x + 7)
Setup: each factor (x - a) gets a term A/(x - a)
Decomposition form: A1/(x - a1) + A2/(x - a2) + ... + An/(x - an)
Some linear factors appear more than once.
Example denominator: Q(x) = x(x - 1)^2 (x + 3)^3
Setup: a factor (x - a)^k gets k terms, one for each power from 1 up to k
Decomposition form for (x - a)^n: A1/(x - a) + A2/(x - a)^2 + ... + An/(x - a)^n
Q(x) has irreducible quadratic factors (complex roots), but each appears only once.
Example denominator: Q(x) = x(x - 1)(x^2 + 4)
Setup: each irreducible quadratic (x^2 + bx + c) gets a term (Ax + B)/(x^2 + bx + c), with a linear numerator rather than just a constant
The linear numerator (Ax + B) is necessary because the denominator is degree 2
The most complex case. Irreducible quadratic factors appear more than once.
Example denominator: Q(x) = x(x - 1)(x^2 + 4)^2
Setup: a repeated quadratic (x^2 + bx + c)^k gets k terms, each with a linear numerator
Decomposition form: (A1 x + B1)/(x^2 + bx + c) + (A2 x + B2)/(x^2 + bx + c)^2 + ... + (Ak x + Bk)/(x^2 + bx + c)^k
Quick-reference for setting up the partial fraction form based on the type of factor in Q(x).
Factor in Q(x) | Decomposition form |
|---|---|
(x - a1)(x - a2)...(x - an) | A1/(x - a1) + A2/(x - a2) + ... + An/(x - an) |
(x - a)^k | A1/(x - a) + A2/(x - a)^2 + ... + Ak/(x - a)^k |
(x^2 + b1 x + c1)...(x^2 + bn x + cn) | (A1 x + B1)/(x^2 + b1 x + c1) + ... + (An x + Bn)/(x^2 + bn x + cn) |
(x^2 + bx + c)^k | (A1 x + B1)/(x^2 + bx + c) + ... + (Ak x + Bk)/(x^2 + bx + c)^k |
Key integration results after decomposition:
A/(x - a) integrates to A ln|x - a| + C
A/(x - a)^n (n > 1) integrates to A/[-(n - 1)(x - a)^(n - 1)] + C
(Ax + B)/(x^2 + c^2) splits into an ln term and an arctan term
Partial fractions show up heavily in engineering and physics. In electrical engineering, they are the standard tool for computing inverse Laplace transforms, which is how you analyse circuit behaviour over time. In control systems, transfer functions are rational functions of a complex variable, and partial fraction decomposition is the first step in determining system stability and response characteristics.
Students often put just a constant A over an irreducible quadratic factor. The numerator must be (Ax + B), a linear expression, not a single constant.
Forgetting to check whether the fraction is proper before starting. If deg(P) >= deg(Q), you must do long division first. Skipping this step gives a wrong decomposition.
When a linear factor repeats, say (x - 1)^3, students sometimes write only one term A/(x - 1). You need three separate terms: one for each power up to the cube.
Mixing up strategic substitution with equating coefficients. Both methods work, but substitution is faster when you have distinct linear factors, while equating coefficients is often necessary for quadratic factors.
Partial fractions is one of the most commonly tested integration techniques in Calc 2. Expect at least one problem on every major exam.
Definite integrals using partial fractions are popular exam questions: you decompose, integrate, then evaluate at the bounds. Watch your signs carefully with ln terms.
Exams frequently combine partial fractions with long division in a single problem, testing whether you check the degree condition first.
The setup step (choosing the correct decomposition form) is where most marks are lost. If your form is wrong, every calculation after it is wrong too.
True or False: You can apply partial fraction decomposition directly to (x^3 + 1)/(x^2 - 1) without any preliminary steps.
Answer: False. The degree of the numerator (3) is greater than the degree of the denominator (2), so you must do long division first.
Fill in the blank: For the factor (x - 5)^2 in the denominator, the decomposition includes ____ separate terms.
Answer: Two. One with denominator (x - 5) and one with denominator (x - 5)^2.
True or False: The partial fraction term for an irreducible quadratic factor (x^2 + 9) has the form A/(x^2 + 9).
Answer: False. It should be (Ax + B)/(x^2 + 9), with a linear numerator.
Fill in the blank: The integral of 3/(x + 2) dx is ______.
Answer: 3 ln|x + 2| + C.
True or False: If Q(x) = (x - 1)(x + 3), you would set up the decomposition as A/(x - 1) + B/(x + 3).
Answer: True. These are distinct linear factors, the simplest case.
Q: Decompose (5x - 4)/(x^2 - x - 2) into partial fractions.
A: Factor the denominator: (x - 2)(x + 1). Set up: A/(x - 2) + B/(x + 1). Multiply through: 5x - 4 = A(x + 1) + B(x - 2). Substituting x = 2 gives A = 2. Substituting x = -1 gives B = 3. Result: 2/(x - 2) + 3/(x + 1).
Q: Evaluate the definite integral of 3/(x^2 - x - 2) dx from 0 to 1.
A: Factor: (x + 1)(x - 2). Decompose: 3/[(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2). Solving gives A = -1, B = 1. Integrate: -ln|x + 1| + ln|x - 2|, evaluated from 0 to 1. Result: (-ln 2 + ln 1) - (-ln 1 + ln 2) = -2 ln 2 = ln(1/4).
Q: Set up (but do not solve) the partial fraction decomposition for 1/[(x - 1)(x^2 + 1)].
A: The decomposition is A/(x - 1) + (Bx + C)/(x^2 + 1). The linear factor gets a constant numerator, the irreducible quadratic gets a linear numerator.
Q: Why must you perform long division on (x^2 - x)/(x^2 + 3x + 5) before decomposing?
A: The degrees of numerator and denominator are equal (both degree 2), so the fraction is not proper. Long division gives 1 + (-4x - 5)/(x^2 + 3x + 5). Only the remainder fraction needs decomposition.
Partial fractions connects directly to polynomial long division (used whenever the fraction is improper) and to factoring techniques from algebra. The integration results rely on your knowledge of ln integrals and inverse trig integrals (particularly arctan). In later courses, this technique is essential for Laplace transforms in differential equations and for analysing transfer functions in signals and systems.
partial fractions, partial fraction decomposition, partial fraction integration, rational function integration, integration techniques Calc 2, M408L, proper fraction, improper fraction, polynomial long division, irreducible quadratic, repeated linear factor, distinct linear factor, Heaviside cover-up method, equating coefficients, strategic substitution, inverse Laplace transform, integral calculus, UT Austin Calc 2