Partial Fractions Integration

The idea

Adding fractions with different denominators creates a single fraction with a complicated denominator: A/(x−1) + B/(x+2) becomes (A(x+2) + B(x−1))/((x−1)(x+2)). Partial fractions runs that arithmetic backwards. Given a rational integrand P(x)/Q(x), we split it back into the simple pieces whose denominators are the factors of Q(x). Each simple piece has a known antiderivative — log or arctan — and the original integral falls out as a sum.

The technique is mechanical once you factor Q(x). The skill is recognizing which decomposition template to use and how to extract the coefficients.

The three decomposition templates

Factor of Q(x)	Contributes	Each piece integrates to
Distinct linear (x − a)	A/(x − a)	A · ln|x − a|
Repeated linear (x − a)ⁿ	A₁/(x − a) + A₂/(x − a)² + ⋯ + Aₙ/(x − a)ⁿ	log (n = 1) or power rule (n ≥ 2)
Irreducible quadratic x² + bx + c	(Bx + C)/(x² + bx + c)	log + arctan after completing the square
Repeated irreducible quadratic (x² + bx + c)ⁿ	Stacked (Bᵢx + Cᵢ)/(x² + bx + c)ⁱ terms	Reduction formulas; rarely seen in practice

"Irreducible" means the quadratic has no real roots — i.e., its discriminant b² − 4c < 0. Reducible quadratics factor into two linear pieces and reduce to the linear cases.

How to apply it

1. Check degree — if deg P ≥ deg Q, do polynomial long division first.
2. Factor Q(x) into linear and irreducible quadratic factors.
3. Write the decomposition template based on the factors.
4. Solve for the unknown constants — by Heaviside cover-up (for distinct linear factors) or by matching coefficients.
5. Integrate each piece — log for simple linear, power rule for repeated, log + arctan for quadratic.
6. Combine. Don't forget + C.

Worked example — ∫ dx / ((x − 1)(x + 2))

The denominator has two distinct linear factors. Template:

1/((x − 1)(x + 2))  =  A/(x − 1)  +  B/(x + 2).

Apply the Heaviside cover-up. To find A, cover up (x − 1) in the original and substitute x = 1:

A  =  1/(x + 2) |_{x = 1}  =  1/(1 + 2)  =  1/3.

B  =  1/(x − 1) |_{x = −2}  =  1/(−2 − 1)  =  −1/3.

Two lines, no algebra. The decomposition is:

1/((x − 1)(x + 2))  =  (1/3)/(x − 1)  −  (1/3)/(x + 2).

Integrate:

∫ dx/((x − 1)(x + 2))
  =  (1/3) ∫ dx/(x − 1)  −  (1/3) ∫ dx/(x + 2)
  =  (1/3) ln|x − 1|  −  (1/3) ln|x + 2|  +  C
  =  (1/3) ln| (x − 1)/(x + 2) |  +  C.

Worked example — ∫ (3x + 1) / (x − 1)² dx (repeated factor)

The denominator has a repeated linear factor (x − 1)². Template:

(3x + 1) / (x − 1)²  =  A/(x − 1)  +  B/(x − 1)².

Clear denominators: 3x + 1 = A(x − 1) + B. Substitute x = 1: 4 = B, so B = 4. Match x-coefficients: 3 = A, so A = 3.

(3x + 1)/(x − 1)²  =  3/(x − 1)  +  4/(x − 1)².

∫ (3x + 1)/(x − 1)² dx  =  3 ln|x − 1|  −  4/(x − 1)  +  C.

The first piece integrates to a log; the second uses the power rule (∫1/(x − 1)² dx = −1/(x − 1) + C).

Worked example — ∫ dx / (x · (x² + 1))

Denominator has one linear factor x and one irreducible quadratic x² + 1 (discriminant −4 < 0). Template:

1/(x(x² + 1))  =  A/x  +  (Bx + C)/(x² + 1).

Clear: 1 = A(x² + 1) + (Bx + C)x = Ax² + A + Bx² + Cx = (A + B)x² + Cx + A.

Match coefficients: x² gives A + B = 0; x gives C = 0; constant gives A = 1. So A = 1, B = −1, C = 0.

1/(x(x² + 1))  =  1/x  −  x/(x² + 1).

∫ dx/(x(x² + 1))  =  ∫ dx/x  −  ∫ x/(x² + 1) dx
                  =  ln|x|  −  (1/2) ln(x² + 1)  +  C
                  =  ln| x / √(x² + 1) |  +  C.

The second integral uses u = x² + 1, du = 2x dx — the (Bx + C)/(x² + 1) piece has a "derivative of denominator" part that goes to log. When C ≠ 0, the remaining 1/(x² + bx + c) piece completes the square and integrates to arctan.

Partial fractions vs other techniques

	Partial fractions	U-substitution	Integration by parts	Trig substitution
Best when	Rational function P(x)/Q(x)	Inner function + derivative	Product of dissimilar factors	Square roots of quadratics
Underlying principle	Algebraic decomposition	Chain rule reversal	Product rule reversal	Pythagorean identity
When to try	For every rational integrand	First — always	Second	When square roots appear
Algebra-heaviness	High (decomposition)	Low	Medium	Medium
Output forms	Logs and arctangents	Anything	uv-style mixed forms	Arctrig and logs
Shortcut tool	Heaviside cover-up	Pattern recognition	LIATE mnemonic	Right-triangle picture

When partial fractions is the right tool

• Any rational integrand. When the integrand is P(x)/Q(x) and u-substitution doesn't apply directly, partial fractions is the systematic next step.
• Inverse Laplace transforms. Most rational Laplace expressions decompose into simple pieces whose inverse transforms are tabulated.
• Partial-fraction expansion in signal processing. Transfer functions H(s) = N(s)/D(s) decompose into modes, each contributing one exponential or sinusoidal response.
• Generating functions in combinatorics. Rational generating functions decompose; each term gives one sequence component.
• Probability — characteristic and moment-generating functions. Decomposition gives separable contributions to distribution computations.
• Differential equations. Method of undetermined coefficients and Laplace methods both lean on partial fractions for the final step.

Common mistakes

• Forgetting to long-divide first. If deg P ≥ deg Q, the decomposition template is wrong. Long-divide to reduce to a polynomial plus a proper rational, then decompose the proper part.
• Wrong template for repeated factors. Repeated (x − a)ⁿ requires n separate terms, one for each power. Using only A/(x − a) gives an underdetermined system.
• Missing the linear numerator for quadratics. Irreducible x² + bx + c needs (Bx + C), not just B. The numerator must span all polynomials of degree < 2.
• Heaviside on the wrong factor type. Cover-up only works for distinct simple linear poles. Repeated factors and quadratics need other methods.
• Sign errors in completing the square. Converting (x² + bx + c) to ((x + b/2)² + (c − b²/4)) requires careful sign tracking.
• Forgetting absolute values in logs. ∫dx/(x − a) = ln|x − a| + C — the absolute value is essential for negative arguments.

Connections to other techniques

U-substitution. After decomposition, some pieces (the derivative-of-denominator part of a quadratic term) reduce by u-sub to logs. Partial fractions sets up u-sub at the term level.

Trig substitution. A Weierstrass substitution t = tan(θ/2) converts rational trig integrals to rational t integrals — partial fractions then finishes them. Trig sub and partial fractions compose.

Laplace transforms. The inverse Laplace transform of a rational H(s) is computed by partial fraction decomposition. The denominator factors give the poles; the residues become the partial-fraction coefficients.

Complex analysis — residues. The Heaviside cover-up is the real-variable face of the residue theorem. A = P(a)/Q'(a) is exactly the residue of P/Q at the simple pole x = a.

Polynomial algebra. Decomposition is finite-dimensional linear algebra in disguise. The space of proper rationals with denominator Q is finite-dimensional, and the partial-fraction basis is a natural decomposition.