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2. Proper Rational Functions0%

Partial Fractions · Topic 2 of 3

2. Proper Rational Functions

Video coming soon4 worked examples

Theory

A rational function is proper when the degree of the numerator is less than the degree of the denominator. Partial fractions reverse the process of adding fractions, splitting one awkward fraction into a sum of simpler ones (which is exactly what you need before integrating). The shape of the decomposition depends on the denominator's factors:

  • Distinct linear factor (x−a)(x-a) gives a term Ax−a\dfrac{A}{x-a}
  • Repeated linear factor (x−a)2(x-a)^2 gives Ax−a+B(x−a)2\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}
  • Irreducible quadratic factor (x2+bx+c)(x^2+bx+c) gives Bx+Cx2+bx+c\dfrac{Bx+C}{x^2+bx+c}

A denominator may contain several of these at once. Each factor contributes its own term (or terms) independently, so (x+1)2(x2+4)(x+1)^2(x^2+4) needs three unknowns in total: two for the repeated linear factor and one linear numerator over the quadratic.

The Golden Rule: factorise the denominator fully and check the fraction is proper before you start. Then choose the correct numerator shape for each factor — a constant over a linear factor, but a linear expression (Bx+C)(Bx+C) over an irreducible quadratic.

⚠️ Common Examiner Traps

  • Wrong numerator over a quadratic: use Bx+CBx+C over an irreducible quadratic, not a single constant.
  • Missing the extra term for a repeated factor: (x−a)2(x-a)^2 needs both Ax−a\dfrac{A}{x-a} and B(x−a)2\dfrac{B}{(x-a)^2}.
  • Method choice: substituting the roots is quickest for linear factors, but you still need to compare coefficients (or substitute an extra value) to pin down the numerator over a quadratic factor.

Worked examples

Example 1

Distinct linear factors. Express 5x−1(x+1)(x−2)\dfrac{5x-1}{(x+1)(x-2)} in partial fractions.

Step 1: Set up the decomposition and clear the denominator:

5x−1(x+1)(x−2)=Ax+1+Bx−2  ⟹  5x−1=A(x−2)+B(x+1)\frac{5x-1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} \implies 5x-1 = A(x-2) + B(x+1)

Step 2: Substitute the roots to find AA and BB:

x=2:  9=3B  ⟹  B=3x=−1: −6=−3A  ⟹  A=2x=2:\ \ 9 = 3B \implies B = 3 \qquad x=-1:\ -6 = -3A \implies A = 2

Step 3: State the result:

5x−1(x+1)(x−2)=2x+1+3x−2\frac{5x-1}{(x+1)(x-2)} = \frac{2}{x+1} + \frac{3}{x-2}

Example 2

Repeated linear factor. Express 3x2−4x+5(x−2)(x−1)2\dfrac{3x^2 - 4x + 5}{(x-2)(x-1)^2} in partial fractions.

Step 1: The repeated factor needs both powers:

3x2−4x+5(x−2)(x−1)2=Ax−2+Bx−1+C(x−1)2\frac{3x^2-4x+5}{(x-2)(x-1)^2} = \frac{A}{x-2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}
3x2−4x+5=A(x−1)2+B(x−2)(x−1)+C(x−2)3x^2-4x+5 = A(x-1)^2 + B(x-2)(x-1) + C(x-2)

Step 2: Substitute the roots, then compare x2x^2 terms for the last unknown:

x=2: 9=Ax=1: 4=−C  ⟹  C=−4x=2:\ 9 = A \qquad x=1:\ 4 = -C \implies C = -4
x2 terms: 3=A+B  ⟹  B=−6x^2\text{ terms}:\ 3 = A + B \implies B = -6

Step 3: State the result:

3x2−4x+5(x−2)(x−1)2=9x−2−6x−1−4(x−1)2\frac{3x^2-4x+5}{(x-2)(x-1)^2} = \frac{9}{x-2} - \frac{6}{x-1} - \frac{4}{(x-1)^2}

Example 3

Irreducible quadratic factor. Express 3x2+1(x+1)(x2+1)\dfrac{3x^2 + 1}{(x+1)(x^2+1)} in partial fractions.

Step 1: Use a linear numerator over the quadratic factor:

3x2+1(x+1)(x2+1)=Ax+1+Bx+Cx2+1\frac{3x^2+1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}
3x2+1=A(x2+1)+(Bx+C)(x+1)3x^2+1 = A(x^2+1) + (Bx+C)(x+1)

Step 2: Substitute the real root, then compare coefficients:

x=−1: 4=2A  ⟹  A=2x=-1:\ 4 = 2A \implies A = 2
x2 terms: 3=A+B  ⟹  B=1constants: 1=A+C  ⟹  C=−1x^2\text{ terms}:\ 3 = A + B \implies B = 1 \qquad \text{constants}:\ 1 = A + C \implies C = -1

Step 3: State the result:

3x2+1(x+1)(x2+1)=2x+1+x−1x2+1\frac{3x^2+1}{(x+1)(x^2+1)} = \frac{2}{x+1} + \frac{x-1}{x^2+1}

Example 4

Repeated linear and irreducible quadratic. Express 2x3+5x2+5x+12(x+1)2(x2+4)\dfrac{2x^3 + 5x^2 + 5x + 12}{(x+1)^2(x^2+4)} in partial fractions.

Step 1: The numerator has degree 33 and the denominator degree 44, so the fraction is proper. Each factor contributes its own terms — four unknowns in all:

2x3+5x2+5x+12(x+1)2(x2+4)=Ax+1+B(x+1)2+Cx+Dx2+4\frac{2x^3+5x^2+5x+12}{(x+1)^2(x^2+4)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx+D}{x^2+4}

Step 2: Multiply through by (x+1)2(x2+4)(x+1)^2(x^2+4):

2x3+5x2+5x+12=A(x+1)(x2+4)+B(x2+4)+(Cx+D)(x+1)22x^3+5x^2+5x+12 = A(x+1)(x^2+4) + B(x^2+4) + (Cx+D)(x+1)^2

Step 3: Substitute the real root x=−1x = -1, which kills the AA and (Cx+D)(Cx+D) terms:

−2+5−5+12=B(1+4)  ⟹  10=5B  ⟹  B=2-2 + 5 - 5 + 12 = B(1+4) \implies 10 = 5B \implies B = 2

Step 4: There are no other real roots, so compare coefficients. Expanding the right-hand side:

x3:A+C=2x2:A+B+2C+D=5constants:4A+4B+D=12\begin{aligned} x^3: &\quad A + C = 2 \\ x^2: &\quad A + B + 2C + D = 5 \\ \text{constants}: &\quad 4A + 4B + D = 12 \end{aligned}

Step 5: Substitute B=2B = 2, giving C=2−AC = 2 - A and D=4−4AD = 4 - 4A. Put both into the x2x^2 equation:

A+2+2(2−A)+(4−4A)=5  ⟹  8−5A=5  ⟹  A=1A + 2 + 2(2-A) + (4-4A) = 5 \implies 8 - 5A = 5 \implies A = 1

Step 6: Back-substitute: C=1C = 1 and D=0D = 0. State the result:

2x3+5x2+5x+12(x+1)2(x2+4)=1x+1+2(x+1)2+xx2+4\frac{2x^3+5x^2+5x+12}{(x+1)^2(x^2+4)} = \frac{1}{x+1} + \frac{2}{(x+1)^2} + \frac{x}{x^2+4}