Partial Fractions

Decompose algebraic fractions into partial fractions at A-Level. Handle linear, repeated, and irreducible factors.

Partial fractions decompose a complex fraction into simpler ones. This is essential for integration and series work at A-Level.

Types

Distinct Linear Factors

f(x)(x−a)(x−b)=Ax−a+Bx−b\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

Repeated Linear Factor

f(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b\frac{f(x)}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}

Improper Fractions

If degree of numerator ≥ degree of denominator, divide first.

Method: Cover-Up (Substitution)

5x+1(x+1)(x−2)\frac{5x + 1}{(x+1)(x-2)}

Let x=−1x = -1: −4−3=43=A\frac{-4}{-3} = \frac{4}{3} = A.

Let x=2x = 2: 113=B\frac{11}{3} = B.

So 5x+1(x+1)(x−2)=4/3x+1+11/3x−2\frac{5x+1}{(x+1)(x-2)} = \frac{4/3}{x+1} + \frac{11/3}{x-2}.

Method: Comparing Coefficients

Equate numerators and compare powers of xx.

Worked Example

3x+5(x+1)(x+2)=Ax+1+Bx+2\frac{3x + 5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}

3x+5=A(x+2)+B(x+1)3x + 5 = A(x+2) + B(x+1)

x=−1x = -1: 2=A(1)2 = A(1) → A=2A = 2. x=−2x = -2: −1=B(−1)-1 = B(-1) → B=1B = 1.

Answer: 2x+1+1x+2\frac{2}{x+1} + \frac{1}{x+2}.

Practice Problems

    1. Express 7x−1(x−1)(x+2)\frac{7x - 1}{(x-1)(x+2)} in partial fractions.
    1. Express 2x2+3(x−1)(x+1)2\frac{2x^2 + 3}{(x-1)(x+1)^2} in partial fractions.
    1. Express x3x2−1\frac{x^3}{x^2 - 1} (improper — divide first).

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Key Takeaways

  • ✓

    Distinct factors: Ax−a+Bx−b\frac{A}{x-a} + \frac{B}{x-b}.

  • ✓

    Repeated: include terms up to the highest power.

  • ✓

    Improper: polynomial division first.

  • ✓

    Use substitution (cover-up) or comparing coefficients.

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