Difficulty: Intermediate | Prerequisites: Calc 1 (basic integration rules), factoring polynomials, long division of polynomials.
Partial fractions is one of the core techniques for integrating rational functions (fractions where both numerator and denominator are polynomials). It sits squarely in the middle of a Calc 2 course, after you have learned basic antiderivative rules and u-substitution, and before you move on to improper integrals and series. The idea is to break a complicated fraction into simpler pieces whose antiderivatives you already know. If you are comfortable factoring quadratics and doing polynomial long division, you have everything you need to start here.
When you need to integrate a fraction like 1/(x² - 1), factor the denominator and rewrite the fraction as a sum of simpler fractions (e.g. A/(x - 1) + B/(x + 1)). Solve for the unknown constants, then integrate each simple fraction separately. The technique also combines with u-substitution, as when a trig integral reduces to a partial-fractions problem.
Rational function
A function of the form P(x)/Q(x) where P and Q are polynomials. In simple terms, any fraction built from polynomials on top and bottom.
Proper rational function
A rational function where the degree of the numerator is strictly less than the degree of the denominator. Think of it as: the top is "smaller" than the bottom, so no polynomial long division is needed first.
Partial fraction decomposition
The process of rewriting a proper rational function as a sum of simpler fractions, each with a denominator that is a factor (or power of a factor) of the original denominator. In simple terms, you are splitting one hard fraction into several easy ones.
Cover-up method (Heaviside method)
A shortcut for finding the constants in a partial fraction decomposition when all denominator factors are distinct and linear. You "cover up" the factor corresponding to a constant, then substitute the root of that factor into what remains. It saves time compared to setting up a full system of equations.
Check the degree. If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first. Work only with the proper remainder.
Factor the denominator completely. Factor into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c where the discriminant is negative).
Write the decomposition form.
Distinct linear factor (x - a): contributes A/(x - a)
Repeated linear factor (x - a)²: contributes A/(x - a) + B/(x - a)²
Irreducible quadratic (x² + bx + c): contributes (Ax + B)/(x² + bx + c)
Solve for the constants. Multiply both sides by the full denominator and either substitute convenient x-values (roots of each factor) or match coefficients of like powers of x.
Integrate each term. Linear denominators give ln|...| terms. Irreducible quadratics give arctan terms (possibly after completing the square).
Factor the denominator: x² - 1 = (x - 1)(x + 1).
Set up the decomposition:
1/(x² - 1) = A/(x - 1) + B/(x + 1)
Multiply both sides by (x - 1)(x + 1):
1 = A(x + 1) + B(x - 1)
Substitute x = 1: 1 = A(2), so A = 1/2.
Substitute x = -1: 1 = B(-2), so B = -1/2.
Therefore:
∫ 1/(x² - 1) dx = (1/2) ∫ 1/(x - 1) dx - (1/2) ∫ 1/(x + 1) dx
= (1/2) ln|x - 1| - (1/2) ln|x + 1| + C
This can also be written as (1/2) ln|(x - 1)/(x + 1)| + C.
The hint from the problem set is to use substitution first. Let u = tan θ, so du = sec²θ dθ.
The integral becomes:
∫ 1/(u² - 1) du
This is exactly the same form as Problem 1. Applying the result:
= (1/2) ln|u - 1| - (1/2) ln|u + 1| + C
= (1/2) ln|tan θ - 1| - (1/2) ln|tan θ + 1| + C
The lesson here: recognise when a substitution transforms a trig integral into a rational function you already know how to decompose.
General decomposition forms (denominator fully factored):
1/((x - a)(x - b)) = A/(x - a) + B/(x - b)
1/((x - a)²) = A/(x - a) + B/(x - a)²
1/((x - a)(x² + bx + c)) = A/(x - a) + (Bx + C)/(x² + bx + c)
Key antiderivatives that result:
∫ 1/(x - a) dx = ln|x - a| + C
∫ 1/(x² + a²) dx = (1/a) arctan(x/a) + C
Partial fractions appear in control systems engineering and electrical engineering when computing inverse Laplace transforms, which is how engineers analyse circuit behaviour and system stability. Any time a transfer function is a ratio of polynomials, partial fractions is the standard tool for breaking it apart.
Students often forget to check whether the rational function is proper before decomposing. If the degree of the numerator is equal to or greater than the denominator, you must do long division first. Skipping this step gives a wrong decomposition.
When the denominator has a repeated factor like (x - 1)², students sometimes write only A/(x - 1) instead of A/(x - 1) + B/(x - 1)². You need one term for each power up to and including the highest.
For irreducible quadratic factors, the numerator in the decomposition must be (Ax + B), not just a constant A. Forgetting the linear term in the numerator leads to an unsolvable system.
Students sometimes drop the absolute value bars in ln|x - a|. This matters when the expression inside could be negative.
⚠️ Partial fractions is a near-certainty on a Calc 2 exam. Expect at least one problem requiring a full decomposition and integration.
⚠️ Problems that combine substitution with partial fractions (like Problem 2 here, where u = tan θ) are a favourite exam pattern. Recognise when a substitution converts a trig integral into a rational function.
⚠️ Make sure you can handle both distinct linear factors and repeated linear factors. Irreducible quadratics are less common at this level but still fair game.
True or false: You can apply partial fractions directly to ∫ (x³ + 1)/(x² - 1) dx without any preliminary step. (Answer: False. The numerator has degree 3, denominator degree 2, so you must do long division first.)
Fill in the blank: 1/((x - 2)(x + 3)) = A/_____ + B/_____. (Answer: A/(x - 2) + B/(x + 3))
True or false: The partial fraction form of 1/(x + 1)² is A/(x + 1). (Answer: False. It should be A/(x + 1) + B/(x + 1)².)
Fill in the blank: After decomposing 1/(x² - 1), the antiderivative involves _____ functions. (Answer: natural logarithm, i.e. ln)
Q: Compute ∫ 1/(x² - 4) dx using partial fractions.
A: Factor: x² - 4 = (x - 2)(x + 2). Decompose: 1/(x² - 4) = A/(x - 2) + B/(x + 2). Solving gives A = 1/4, B = -1/4. Result: (1/4) ln|x - 2| - (1/4) ln|x + 2| + C.
Q: Compute ∫ sec²θ / (tan²θ - 1) dθ.
A: Substitute u = tan θ, du = sec²θ dθ. The integral becomes ∫ 1/(u² - 1) du = (1/2) ln|u - 1| - (1/2) ln|u + 1| + C = (1/2) ln|tan θ - 1| - (1/2) ln|tan θ + 1| + C.
Q: What is the first step when asked to find ∫ (x² + 3)/(x² - 1) dx?
A: Polynomial long division, because the degree of the numerator equals the degree of the denominator. Dividing gives 1 + 4/(x² - 1), and you then decompose 4/(x² - 1) using partial fractions.
Q: Decompose 3/((x - 1)(x + 1)²) into partial fractions. What form do you use?
A: The correct form is A/(x - 1) + B/(x + 1) + C/(x + 1)². The repeated factor (x + 1)² requires two terms.
Partial fractions connects directly to u-substitution (Problem 2 is a perfect example of chaining the two techniques). It also appears again when you study improper integrals, since many convergence problems involve rational functions. In differential equations, partial fractions is essential for solving separable ODEs and for computing inverse Laplace transforms.
Partial fraction decomposition, partial fractions integration, rational function integration, Heaviside cover-up method, integrating 1/(x²-1), polynomial long division for integrals, Calc 2 antiderivatives, NYU Calculus II PS2, sec²θ/(tan²θ-1) integral, u-substitution with partial fractions