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Partial Fractions and Integration

Key Terms

Purpose
A rational function P(x)/Q(x) is split into simpler fractions whose integrals are standard forms (ln|...| or arctan(...)).
Proper fraction
deg(P) < deg(Q). If not proper, divide first to get a polynomial plus a proper fraction.
Linear factors
For each factor (ax + b) in Q(x), include a term A/(ax + b) in the decomposition.
Repeated linear factor
For (ax + b)², include A/(ax+b) + B/(ax+b)². For (ax+b)n, include n terms.
Irreducible quadratic factor
For (x² + bx + c) that doesn’t factorise over ℜ, include (Ax + B)/(x² + bx + c).
Finding constants
Multiply both sides by Q(x) to clear denominators. Then either substitute convenient values of x (cover-up method), or compare coefficients.
Integration
∫A/(ax+b) dx = (A/a)ln|ax+b| + C.   ∫(Ax+B)/(x²+k²) dx splits into a log term and arctan term.
Standard Forms After Partial Fractions:
  • ∫ A/(ax + b) dx = (A/a) ln|ax + b| + C
  • ∫ A/(ax + b)² dx = −A/(a(ax + b)) + C
  • ∫ 1/(x² + k²) dx = (1/k) arctan(x/k) + C
  • ∫ x/(x² + k²) dx = (1/2) ln(x² + k²) + C
Hot Tip The cover-up method for finding constants: to find A in A/(x−a), mentally cover up (x−a) in the original denominator and substitute x = a into what remains. This gives A directly for distinct linear factors. For repeated or quadratic factors, you must use coefficient comparison or substitute multiple values.

Worked Example 1 — Distinct linear factors

Find ∫ 1/((x+1)(x−2)) dx.

Write 1/((x+1)(x−2)) = A/(x+1) + B/(x−2). Multiply through by (x+1)(x−2):

1 = A(x−2) + B(x+1).

Set x = −1: 1 = A(−3) ⇒ A = −1/3.   Set x = 2: 1 = B(3) ⇒ B = 1/3.

∫ dx/((x+1)(x−2)) = ∫[−1/(3(x+1)) + 1/(3(x−2))] dx

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

Worked Example 2 — Repeated linear factor

Find ∫ (x+3)/(x+1)² dx.

Write (x+3)/(x+1)² = A/(x+1) + B/(x+1)². Multiply through:

x+3 = A(x+1) + B. Set x = −1: 2 = B. Compare x-coefficients: 1 = A.

∫ (x+3)/(x+1)² dx = ∫ [1/(x+1) + 2/(x+1)²] dx

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

Why Partial Fractions Work

A rational function like 5/((x+1)(x+3)) looks complicated, but it is actually the sum of simpler fractions: 2/(x+3) − 2/(x+1)... wait, let’s work it out properly. The key insight is that any rational function whose denominator factors completely over the reals can be broken into pieces where the denominator is a power of a linear or irreducible quadratic. These pieces integrate using only logarithms and arctangents — standard forms that appear on the QCAA formula sheet.

The technique works by reversing the process of adding fractions. When you add A/(x−1) + B/(x+2), you get a rational function with denominator (x−1)(x+2). Partial fractions goes in the other direction: given the combined fraction, find A and B.

Step-by-Step Procedure

  1. Check if proper: Is deg(numerator) < deg(denominator)? If not, perform polynomial long division first. Write the result as a polynomial + proper fraction.
  2. Factorise the denominator completely over ℜ.
  3. Write the partial fraction form based on the factor types (see Key Ideas).
  4. Find the constants by multiplying both sides by the denominator and using substitution or coefficient comparison.
  5. Integrate each partial fraction term separately.

Handling an Improper Fraction

If deg(numerator) ≥ deg(denominator), perform long division. For example, (x³ + 1)/(x² − 1): dividing x³ + 1 by x² − 1 gives quotient x with remainder x + 1. So (x³+1)/(x²−1) = x + (x+1)/(x²−1). Now apply partial fractions to the proper part (x+1)/((x−1)(x+1)) = 1/(x−1).

Irreducible Quadratic Factors

For a factor x² + k² (or completing the square for x² + bx + c with negative discriminant), the numerator of the partial fraction term is Ax + B, not just a constant. This is because the degree of the numerator must be less than the degree of that denominator factor. To integrate (Ax + B)/(x² + k²), split it:

  • ∫ Ax/(x² + k²) dx = (A/2)ln(x² + k²) + C   (substitution u = x²+k²)
  • ∫ B/(x² + k²) dx = (B/k) arctan(x/k) + C

The Cover-Up Method

For distinct linear factors, the constant A for the term A/(x−a) can be found instantly: multiply the original fraction by (x−a) and set x = a. For example, in 1/((x−1)(x+2)) = A/(x−1) + B/(x+2):

  • A = [cover (x−1)]: evaluate 1/(x+2) at x = 1 ⇒ A = 1/3.
  • B = [cover (x+2)]: evaluate 1/(x−1) at x = −2 ⇒ B = −1/3.

This shortcut only applies to distinct linear factors; for repeated factors or quadratics, fall back to multiplying out and comparing coefficients.

Exam Tip: After finding your constants, do a quick sanity check by recombining the partial fractions over the common denominator. If you get back the original numerator, your constants are correct. This takes 30 seconds and prevents losing marks on the integration steps.
Exam Tip: When a definite integral involves partial fractions, you can integrate first (leaving the answer in terms of ln and arctan), then apply the limits. Do not try to re-combine the logarithms before substituting — it introduces unnecessary complexity.

Mastery Practice

  1. Find ∫ 1/((x+1)(x+3)) dx. Fluency

  2. Find ∫ (x+5)/((x−1)(x+2)) dx. Fluency

  3. Find ∫ (3x² + x + 2)/(x(x² + 1)) dx. Fluency

  4. Find ∫ 1/(x² − 4) dx. Fluency

  5. Find ∫ (2x+3)/(x+1)² dx using a repeated linear factor. Understanding

  6. Evaluate ∫12 (x+1)/(x(x−1)) dx. (Hint: note the factor x−1 in the denominator, which has a zero at x=1 outside the interval.) Understanding

  7. Find ∫ (x² + 2)/((x² + 1)(x + 1)) dx. Understanding

  8. Find ∫ (4x² − 3x + 2)/(x²(x − 1)) dx. Understanding

  9. Find ∫ (x³ + 1)/(x² − 1) dx. (This is an improper fraction — perform long division first.) Problem Solving

  10. Use partial fractions to evaluate ∫01/2 1/(1−x²) dx and verify using the identity 1/(1−x²) = (1/2)[1/(1−x) + 1/(1+x)]. Problem Solving

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