Difficulty: Intermediate | Prerequisites: Calc 1 (basic integration, u-substitution), comfort with algebraic manipulation and factoring polynomials.
When you try to integrate a rational function (a polynomial divided by another polynomial), the first question is whether the numerator's degree is at least as large as the denominator's. If it is, you cannot jump straight to partial fractions. You need to do polynomial long division first to rewrite the integrand as a polynomial plus a proper fraction, and only then decompose the leftover fraction.
This technique sits right at the start of the "integration of rational functions" unit in most Calc 2 courses. It is a prerequisite for every partial-fractions problem where the rational function is improper. If you are comfortable with algebraic long division from secondary school, the mechanics here are the same, just applied to polynomials rather than numbers.
If the degree of the numerator is greater than or equal to the degree of the denominator, divide first. The result is a polynomial plus a proper fraction. Then decompose that proper fraction into partial fractions and integrate each piece separately.
Rational function
A function of the form P(x)/Q(x), where P and Q are both polynomials. In simple terms, it is one polynomial sitting on top of another in a fraction.
Improper rational function
A rational function where deg P(x) ≥ deg Q(x), meaning the numerator's degree is at least as large as the denominator's. Think of it as the "top-heavy" case: the fraction cannot be decomposed into partial fractions until you deal with the extra degree first.
Proper rational function
A rational function where deg P(x) < deg Q(x). This is the form you need before applying partial fraction decomposition. In simple terms, the bottom is bigger than the top.
Polynomial long division
The algebraic process of dividing one polynomial by another, producing a polynomial quotient and a remainder. It works exactly like the long division you learned with numbers, but with polynomial terms instead of digits.
Partial fraction decomposition
The technique of breaking a proper rational function into a sum of simpler fractions, each with a linear or irreducible quadratic denominator. Think of it as reverse-engineering the addition of fractions: you are splitting one complicated fraction back into the simple pieces that were added together to make it.
Check degrees first: if deg P(x) ≥ deg Q(x), you must divide before doing anything else.
The division rewrites P(x)/Q(x) as S(x) + R(x)/Q(x), where S(x) is the polynomial quotient and R(x) is the remainder with deg R < deg Q.
Once you have the proper fraction R(x)/Q(x), apply partial fractions to that piece only.
Arrange both polynomials in descending powers of x. Fill in any missing terms with a zero coefficient.
Divide the leading term of the numerator by the leading term of the denominator. Write this as the first term of the quotient.
Multiply the entire denominator by that term and subtract the result from the numerator.
Bring down the next term and repeat until the degree of the remainder is strictly less than the degree of the denominator.
Write the result as: quotient + remainder/denominator.
Simplify (x² + 2x − 7) / (x − 2).
Divide x² by x to get x. Multiply (x − 2) by x to get x² − 2x. Subtract: (x² + 2x − 7) − (x² − 2x) = 4x − 7.
Divide 4x by x to get 4. Multiply (x − 2) by 4 to get 4x − 8. Subtract: (4x − 7) − (4x − 8) = 1.
Result: (x² + 2x − 7)/(x − 2) = x + 4 + 1/(x − 2).
Verification: (x − 2)(x + 4) + 1 = x² + 2x − 7. ✓
Evaluate ∫ (x² + 1)/(x² − 1) dx.
Degrees are equal (both degree 2), so divide first.
x² + 1 divided by x² − 1 gives quotient 1 with remainder 2. So the integrand becomes 1 + 2/(x² − 1).
Factor the denominator: x² − 1 = (x − 1)(x + 1).
Set up partial fractions: 2/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1).
Multiply through: 2 = A(x + 1) + B(x − 1).
Matching coefficients: A + B = 0 and A − B = 2, giving A = 1, B = −1.
The integral becomes ∫ [1 + 1/(x − 1) − 1/(x + 1)] dx = x + ln|x − 1| − ln|x + 1| + C.
Evaluate ∫ (3x + 1)/(x² − x − 6) dx.
The numerator is degree 1, denominator is degree 2, so this is already a proper fraction. No long division needed.
Factor: x² − x − 6 = (x + 2)(x − 3).
Partial fractions: (3x + 1)/((x + 2)(x − 3)) = A/(x + 2) + B/(x − 3).
Multiply through: 3x + 1 = A(x − 3) + B(x + 2).
Substituting x = 3: 10 = 5B, so B = 2. Substituting x = −2: −5 = −5A, so A = 1.
Result: ∫ [1/(x + 2) + 2/(x − 3)] dx = ln|x + 2| + 2 ln|x − 3| + C.
Evaluate ∫ (x⁴ − 5x³ + 6x² − 18)/(x³ − 3x²) dx.
Degree 4 over degree 3: long division is needed.
Dividing gives quotient (x − 2) with remainder (6x² − 18) over (x³ − 3x²) remaining. So the integrand is x − 2 + (6x² − 18)/(x³ − 3x²).
Factor the denominator: x³ − 3x² = x²(x − 3).
Partial fractions for 18/(x²(x − 3)): set 18/(x²(x − 3)) = A/x + B/x² + C/(x − 3).
Note: x² is a repeated linear factor, so it needs both A/x and B/x² terms.
Solving: 18 = Ax(x − 3) + B(x − 3) + Cx². Substituting x = 0 gives B = −6. Substituting x = 3 gives C = 2. Expanding and matching x² coefficients gives A + C = 0, so A = −2.
Result: ∫ [x − 2 + (−2)/x + (−6)/x² + 2/(x − 3)] dx = x²/2 − 2x − 2 ln|x| + 6/x + 2 ln|x − 3| + C.
\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}, \quad \deg R < \deg QFor distinct linear factors in the denominator:
\frac{R(x)}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}For a repeated linear factor:
\frac{R(x)}{(x - a)^2 (x - b)} = \frac{A}{x - a} + \frac{B}{(x - a)^2} + \frac{C}{x - b}Key integration results used throughout:
∫ 1/(x − a) dx = ln|x − a| + C
∫ 1/(x − a)² dx = −1/(x − a) + C
Partial fraction decomposition appears in engineering whenever you need to find the inverse Laplace transform of a transfer function, which is how control systems engineers analyse circuit response and mechanical vibrations. It also shows up in probability when deriving the distribution of sums of exponential random variables.
Students often try to apply partial fractions directly to an improper rational function without dividing first. This does not work. You must divide until the remainder's degree is strictly less than the denominator's degree.
When the denominator has a repeated factor like x², students frequently set up only one fraction (A/x²) instead of the required two (A/x + B/x²). Every power of the repeated factor up to and including the highest needs its own term.
A common algebra error is forgetting to distribute the negative sign when subtracting during long division. This cascades through the rest of the calculation and produces a wrong remainder.
Students sometimes confuse the method for finding A and B. You can either substitute convenient values of x (the "cover-up" method) or expand and match coefficients. Both work; pick whichever feels more natural for the problem.
⚠️ The first thing to check on any rational-function integration problem is whether long division is needed. Exams routinely test whether you know to check degrees before setting up partial fractions.
⚠️ Repeated linear factors (e.g. x²) require multiple partial-fraction terms. This is a favourite exam trap.
⚠️ Always verify your partial-fraction constants by plugging them back into the original equation. A sign error in A or B produces a wrong antiderivative that is difficult to catch later.
⚠️ Be comfortable with both methods for solving for constants: substituting roots of the denominator factors and expanding/matching coefficients. Some problems lend themselves to one method more than the other.
True or false: You can apply partial fraction decomposition to (x³ + 1)/(x² − 4) without doing long division first.
False. deg 3 ≥ deg 2, so you must divide first.
Fill in the blank: When the denominator contains (x − 1)², the partial fraction setup needs terms with denominators ______ and ______.
(x − 1) and (x − 1)².
True or false: (x + 3)/((x − 2)(x + 5)) = A/(x − 2) + B/(x + 5) is the correct partial-fraction form.
True. Two distinct linear factors, one constant per factor.
Fill in the blank: In polynomial long division, you stop dividing when the degree of the remainder is ______ the degree of the divisor.
Strictly less than.
Q: Evaluate ∫ (x² + 2x − 7)/(x − 2) dx.
A: Divide first to get x + 4 + 1/(x − 2). Integrate term by term: x²/2 + 4x + ln|x − 2| + C.
Q: Set up the partial fraction decomposition for 2/(x² − 1). Do not solve for the constants.
A: Factor: x² − 1 = (x − 1)(x + 1). Decomposition: 2/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1).
Q: Evaluate ∫ (3x + 1)/(x² − x − 6) dx.
A: Factor: (x + 2)(x − 3). Partial fractions give A = 1, B = 2. Integral: ln|x + 2| + 2 ln|x − 3| + C.
Q: Why does the partial-fraction setup for 18/(x²(x − 3)) require three terms rather than two?
A: Because x² is a repeated linear factor. You need A/x and B/x² to cover both powers of the repeated root, plus C/(x − 3) for the distinct factor.
Q: What is the first step when asked to integrate P(x)/Q(x) and deg P ≥ deg Q?
A: Perform polynomial long division to rewrite the integrand as a polynomial plus a proper rational function.
This material connects directly to integration by partial fractions with irreducible quadratic factors, which extends the technique to denominators that do not factor into linear terms (e.g. x² + 1). You will use the same long-division step there.
It also underpins the method of integrating rational functions that appear after trigonometric substitution or Weierstrass substitution, where the result of the substitution is often a rational function that needs decomposition.
In differential equations (typically the next course), partial fractions appear constantly when solving linear ODEs with the Laplace transform.
Polynomial long division, synthetic division, partial fraction decomposition, partial fractions integration, improper rational function, proper rational function, rational function integration, Calc 2 integration techniques, integration of rational functions, distinct linear factors, repeated linear factors, cover-up method, Heaviside method, UT Austin Calc 2, M408D, integral calculus