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Integration — The use of partial fractions in integration

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Use of Partial Fractions (9709)

Partial fractions rewrite a complicated rational expression into simpler fractions that can be integrated or simplified easily.

1. What Are Partial Fractions?

A rational expression of the form \(\dfrac{P(x)}{Q(x)}\) can be rewritten as a sum of simpler fractions, provided the degree of the numerator is less than the degree of the denominator.

Example factorizations:

  • \(\dfrac{1}{x(x+1)} = \dfrac{A}{x} + \dfrac{B}{x+1}\)
  • \(\dfrac{3x+1}{(x-2)(x+4)} = \dfrac{A}{x-2} + \dfrac{B}{x+4}\)

2. Forms You Must Learn

  • Two distinct linear factors:
    \[ \frac{A}{(x-a)} + \frac{B}{(x-b)} \]
  • Repeated linear factors:
    \[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} \]
  • Irreducible quadratic factor:
    \[ \frac{Ax + B}{x^2 + px + q} \]
  • Improper fraction (numerator degree ≥ denominator):
    FIRST divide: \[ \frac{2x^2 + 5x + 3}{x+1} = 2x + 3 + \frac{0}{x+1} \]

3. Common 9709 Examples

Example 1: \(\displaystyle \frac{5x+3}{x(x+2)}\)

Assume \[ \frac{5x+3}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2} \] Multiply by \(x(x+2)\): \[ 5x + 3 = A(x+2) + Bx \] Expand: \[ 5x + 3 = (A + B)x + 2A \] Compare coefficients: \[ A + B = 5,\qquad 2A = 3 \] So \(A = \tfrac{3}{2}\), \(B = \tfrac{7}{2}\).


Example 2 (quadratic denominator): \(\displaystyle \frac{2x+7}{(x+1)(x^2+4)}\)

Assume \[ \frac{2x+7}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 4} \] Multiply by denominator: \[ 2x + 7 = A(x^2 + 4) + (Bx + C)(x+1) \] Expand and compare coefficients to solve for \(A, B, C\).


Example 3 (Repeated linear): \(\displaystyle \frac{4}{x^2(x+3)}\)

Form: \[ \frac{4}{x^2(x+3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+3} \] Multiply by \(x^2(x+3)\), expand, solve A, B, C.


Example 4 (Improper fraction): \(\displaystyle \frac{2x^2 + 3x + 1}{x+2}\)

First divide: \[ \frac{2x^2 + 3x + 1}{x+2} = 2x - 1 + \frac{3}{x+2} \] Now it's ready for integration.


Example 5 (Integration): \(\displaystyle \int \frac{6x+5}{x^2 - x} \, dx\)

Factor denominator: \[ x^2 - x = x(x-1) \] Write: \[ \frac{6x + 5}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1} \] Solve for \(A\), \(B\).

Then integrate: \[ \int \frac{A}{x}\,dx = A\ln|x|,\qquad \int \frac{B}{x-1}\,dx = B\ln|x-1| \]

4. Integrals You MUST Know

Once split into partial fractions, you typically use these:

  • \(\displaystyle \int \frac{1}{x} dx = \ln|x|\)
  • \(\displaystyle \int \frac{1}{x-a} dx = \ln|x-a|\)
  • \(\displaystyle \int \frac{x}{x^2 + a^2} dx = \tfrac{1}{2}\ln(x^2 + a^2)\)
  • \(\displaystyle \int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right)\)
  • \(\displaystyle \int \frac{Ax + B}{x^2 + a^2} dx = A\cdot\frac{1}{2}\ln(x^2 + a^2) + B\cdot\frac{1}{a}\arctan\!\left(\frac{x}{a}\right)\)

5. 9709 Exam Tips

  • Always check if the fraction is proper. If not, divide first.
  • Always factor the denominator correctly.
  • Use the correct form for repeated or quadratic factors.
  • After multiplying by the denominator, expand neatly and compare coefficients.
  • Signs matter—very common source of lost marks.
  • Integration at the end is usually straightforward (logs + arctan).
  • For definite integrals, substitute limits ONLY at the end.
  • Show working: Cambridge awards method marks for correct set-up.
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