A rational function is a fraction of the form \[ \frac{P(x)}{Q(x)}, \] where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) \ne 0\).

• It is proper if \(\deg P < \deg Q\).
• It is improper if \(\deg P \ge \deg Q\).

In integration we often rewrite a rational function as a sum of simpler fractions — this is called partial fraction decomposition.

Common patterns

• Distinct linear factors: \[ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}. \]
• Repeated linear factor: \[ \frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}. \]
• Linear and repeated linear: \[ \frac{P(x)}{(x-a)(x-b)^2} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{(x-b)^2}. \]
• Irreducible quadratic factor: \[ \frac{P(x)}{(x-a)(x^2+px+q)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+px+q}. \]

Basic method (for proper fractions)

1. Factorise the denominator completely.
2. Write down the correct partial fraction form using unknowns \(A,B,C,\dots\).
3. Multiply both sides by the denominator to remove fractions.
4. Either:
   • substitute convenient values of \(x\), and/or
   • expand and equate coefficients for each power of \(x\).
5. Solve for the unknown constants.

Question 1

Express each of the following as partial fractions.

(a) \(\displaystyle \frac{6x - 2}{(x - 2)(x + 3)}\)

Assume \[ \frac{6x - 2}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}. \] Then \[ 6x - 2 = A(x+3) + B(x-2). \] Putting \(x = 2\): \(12 - 2 = A(5)\Rightarrow A = 2\).
Putting \(x = -3\): \(-18 - 2 = B(-5)\Rightarrow B = 4\).

\[ \boxed{\dfrac{6x - 2}{(x - 2)(x + 3)} = \frac{2}{x - 2} + \frac{4}{x + 3}}. \]

(b) \(\displaystyle \frac{7x + 12}{2x(x - 4)}\)

\[ \frac{7x + 12}{2x(x - 4)} = \frac{A}{x} + \frac{B}{x - 4}. \] So \[ 7x + 12 = A(x - 4)\cdot 2 + B(2x) = 2A(x-4) + 2Bx. \] Put \(x = 0\): \(12 = 2A(-4)\Rightarrow A = -\dfrac{3}{2}\).
Put \(x = 4\): \(28 + 12 = 2B\cdot4\Rightarrow B = 5\).

\[ \boxed{\dfrac{7x + 12}{2x(x - 4)} = -\frac{3}{2x} + \frac{5}{x - 4}} \;=\; \boxed{\frac{5}{x - 4} - \frac{3}{2x}}. \]

(c) \(\displaystyle \frac{15x + 13}{(x - 1)(3x + 1)}\)

\[ \frac{15x + 13}{(x - 1)(3x + 1)} = \frac{A}{x - 1} + \frac{B}{3x + 1}. \] Then \[ 15x + 13 = A(3x+1) + B(x-1). \] Put \(x = 1\): \(15+13 = A(4)\Rightarrow A = 7\).
Put \(x = -\frac{1}{3}\): \(15\left(-\frac{1}{3}\right)+13 = B\left(-\frac{1}{3}-1\right)\Rightarrow B = -6\).

\[ \boxed{\dfrac{15x + 13}{(x - 1)(3x + 1)} = \frac{7}{x - 1} - \frac{6}{3x + 1}}. \]

(d) \(\displaystyle \frac{x - 1}{(3x - 5)(x - 3)}\)

\[ \frac{x - 1}{(3x - 5)(x - 3)} = \frac{A}{3x - 5} + \frac{B}{x - 3}. \] So \[ x - 1 = A(x-3) + B(3x-5). \] Put \(x = 3\): \(3 - 1 = B(9-5)\Rightarrow B = \frac{1}{2}\).
Put \(x = \frac{5}{3}\): \(\frac{5}{3}-1 = A\left(\frac{5}{3}-3\right)\Rightarrow A = -\frac{1}{2}\).

\[ \boxed{\dfrac{x - 1}{(3x - 5)(x - 3)} = -\frac{1}{2(3x - 5)} + \frac{1}{2(x - 3)}}. \]

(e) \(\displaystyle \frac{6x^{2} + 5x - 2}{x(x - 1)(2x + 1)}\)

\[ \frac{6x^{2} + 5x - 2}{x(x - 1)(2x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{2x + 1}. \] Multiplying through and solving (by substitution or equating coefficients) gives \[ A = 2,\quad B = 3,\quad C = -4. \]

\[ \boxed{\dfrac{6x^{2} + 5x - 2}{x(x - 1)(2x + 1)} = \frac{2}{x} + \frac{3}{x - 1} - \frac{4}{2x + 1}}. \]

(f) \(\displaystyle \frac{11x + 12}{(2x + 3)(x + 2)(x - 3)}\)

\[ \frac{11x + 12}{(2x + 3)(x + 2)(x - 3)} = \frac{A}{2x + 3} + \frac{B}{x + 2} + \frac{C}{x - 3}. \] Solving for \(A,B,C\) gives \[ A = 2,\quad B = -2,\quad C = 1. \]

\[ \boxed{\dfrac{11x + 12}{(2x + 3)(x + 2)(x - 3)} = \frac{2}{2x + 3} - \frac{2}{x + 2} + \frac{1}{x - 3}}. \]

Question 2

Express each of the following as partial fractions.

(a) \(\displaystyle \frac{2x}{(x + 2)^{2}}\)

\[ \frac{2x}{(x + 2)^{2}} = \frac{A}{x + 2} + \frac{B}{(x + 2)^{2}}. \] Then \[ 2x = A(x+2) + B. \] Put \(x = -2\): \(2(-2) = B\Rightarrow B = -4\).
Compare coefficients of \(x\): \(2 = A\Rightarrow A = 2\).

\[ \boxed{\dfrac{2x}{(x + 2)^{2}} = \frac{2}{x + 2} - \frac{4}{(x + 2)^{2}}}. \]

(b) \(\displaystyle \frac{11x^{2} + 14x + 5}{(2x + 1)(x + 1)^{2}}\)

\[ \frac{11x^{2} + 14x + 5}{(2x + 1)(x + 1)^{2}} = \frac{A}{2x + 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}. \] Solving for \(A,B,C\) gives \[ A = 3,\quad B = 4,\quad C = -2. \]

\[ \boxed{\dfrac{11x^{2} + 14x + 5}{(2x + 1)(x + 1)^{2}} = \frac{3}{2x + 1} + \frac{4}{x + 1} - \frac{2}{(x + 1)^{2}}}. \]

(c) \(\displaystyle \frac{x^{2} - 2}{x(x - 1)^{2}}\)

\[ \frac{x^{2} - 2}{x(x - 1)^{2}} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^{2}}. \] Solving gives \[ A = -2,\quad B = 3,\quad C = -1. \]

\[ \boxed{\dfrac{x^{2} - 2}{x(x - 1)^{2}} = -\frac{2}{x} + \frac{3}{x - 1} - \frac{1}{(x - 1)^{2}}}. \]

(d) \(\displaystyle \frac{36x^{2} + 2x - 4}{(2x - 3)(2x + 1)^{2}}\)

\[ \frac{36x^{2} + 2x - 4}{(2x - 3)(2x + 1)^{2}} = \frac{A}{2x - 3} + \frac{B}{2x + 1} + \frac{C}{(2x + 1)^{2}}. \] Solving gives \[ A = 5,\quad B = 4,\quad C = -1. \]

\[ \boxed{\dfrac{36x^{2} + 2x - 4}{(2x - 3)(2x + 1)^{2}} = \frac{5}{2x - 3} + \frac{4}{2x + 1} - \frac{1}{(2x + 1)^{2}}}. \]

(e) \(\displaystyle \frac{3}{(x + 2)(x - 2)^{2}}\)

\[ \frac{3}{(x + 2)(x - 2)^{2}} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C}{(x - 2)^{2}}. \] Solving gives \[ A = \frac{3}{16},\quad B = -\frac{3}{16},\quad C = \frac{3}{4}. \]

\[ \boxed{\dfrac{3}{(x + 2)(x - 2)^{2}} = \frac{3}{16(x + 2)} - \frac{3}{16(x - 2)} + \frac{3}{4(x - 2)^{2}}}. \]

(f) \(\displaystyle \frac{3x + 4}{(x + 2)(x - 1)^{2}}\)

\[ \frac{3x + 4}{(x + 2)(x - 1)^{2}} = \frac{A}{x + 2} + \frac{B}{x - 1} + \frac{C}{(x - 1)^{2}}. \] Solving gives \[ A = -\frac{2}{9},\quad B = \frac{2}{9},\quad C = \frac{7}{3}. \]

\[ \boxed{\dfrac{3x + 4}{(x + 2)(x - 1)^{2}} = -\frac{2}{9(x + 2)} + \frac{2}{9(x - 1)} + \frac{7}{3(x - 1)^{2}}}. \]

Question 3

Express each of the following as partial fractions.

(a) \(\displaystyle \frac{2x^{2} - 3x + 2}{x(x^{2} + 1)}\)

\[ \frac{2x^{2} - 3x + 2}{x(x^{2} + 1)} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}. \] Solving gives \(A = 2,\; B = 0,\; C = -3\).

\[ \boxed{\dfrac{2x^{2} - 3x + 2}{x(x^{2} + 1)} = \frac{2}{x} - \frac{3}{x^{2} + 1}}. \]

(b) \(\displaystyle \frac{3x^{2} + 4x + 17}{(2x + 1)(x^{2} + 5)}\)

\[ \frac{3x^{2} + 4x + 17}{(2x + 1)(x^{2} + 5)} = \frac{A}{2x + 1} + \frac{Bx + C}{x^{2} + 5}. \] Solving gives \(A = 3,\; B = 0,\; C = 2\).

\[ \boxed{\dfrac{3x^{2} + 4x + 17}{(2x + 1)(x^{2} + 5)} = \frac{3}{2x + 1} + \frac{2}{x^{2} + 5}}. \]

(c) \(\displaystyle \frac{2x^{2} - 6x - 9}{(3x + 5)(2x^{2} + 1)}\)

\[ \frac{2x^{2} - 6x - 9}{(3x + 5)(2x^{2} + 1)} = \frac{A}{3x + 5} + \frac{Bx + C}{2x^{2} + 1}. \] Solving gives \(A = 1,\; B = 0,\; C = -2\).

\[ \boxed{\dfrac{2x^{2} - 6x - 9}{(3x + 5)(2x^{2} + 1)} = \frac{1}{3x + 5} - \frac{2}{2x^{2} + 1}}. \]

(d) \(\displaystyle -\frac{6x^{2} - 21x + 50}{(3x - 5)(2x^{2} + 5)}\)

\[ -\frac{6x^{2} - 21x + 50}{(3x - 5)(2x^{2} + 5)} = \frac{A}{3x - 5} + \frac{Bx + C}{2x^{2} + 5}. \] Solving gives \(A = -3,\; B = 0,\; C = 7\).

\[ \boxed{-\dfrac{6x^{2} - 21x + 50}{(3x - 5)(2x^{2} + 5)} = -\frac{3}{3x - 5} + \frac{7}{2x^{2} + 5}} \;=\; \boxed{\frac{7}{2x^{2} + 5} - \frac{3}{3x - 5}}. \]

For an improper fraction, first carry out polynomial division to write it as \[ \text{polynomial} + \text{proper fraction}, \] then decompose the proper fraction as before.

Question 4

(a) \(\displaystyle \frac{2x^{2} + 3x + 4}{(x - 1)(x + 2)}\)

After division and decomposition: \[ \boxed{\dfrac{2x^{2} + 3x + 4}{(x - 1)(x + 2)} = 2 + \frac{3}{x - 1} - \frac{2}{x + 2}}. \]

(b) \(\displaystyle \frac{x^{2} + 3}{x^{2} - 4}\)

Denominator: \(x^{2} - 4 = (x - 2)(x + 2)\). Then \[ \boxed{\dfrac{x^{2} + 3}{x^{2} - 4} = 1 + \frac{7}{4(x - 2)} - \frac{7}{4(x + 2)}} \;=\; \boxed{1 - \frac{7}{4(x + 2)} + \frac{7}{4(x - 2)}}. \]

(c) \(\displaystyle \frac{22 - 17x + 21x^{2} - 4x^{3}}{(x - 4)(x^{2} + 1)}\)

After division and decomposition: \[ \boxed{\dfrac{22 - 17x + 21x^{2} - 4x^{3}}{(x - 4)(x^{2} + 1)} = -4 + \frac{2}{x - 4} + \frac{3x - 1}{x^{2} + 1}}. \]

(d) \(\displaystyle \frac{4x^{3} + x^{2} - 16x + 7}{2x^{3} - 4x^{2} + 2x}\)

Note \(2x^{3} - 4x^{2} + 2x = 2x(x - 1)^{2}\). Then \[ \boxed{\dfrac{4x^{3} + x^{2} - 16x + 7}{2x^{3} - 4x^{2} + 2x} = 2 + \frac{7}{2x} + \frac{1}{x - 1} - \frac{2}{(x - 1)^{2}}}. \]

Question 5

Find the values of \(A,B,C,D\) such that \[ \frac{4x^{3} - 9x^{2} + 11x - 4}{x^{2}(2x - 1)} \equiv A + \frac{B}{x} + \frac{C}{x^{2}} + \frac{D}{2x - 1}. \]

Solution.

Multiply both sides by \(x^{2}(2x - 1)\): \[ 4x^{3} - 9x^{2} + 11x - 4 = A x^{2}(2x - 1) + Bx(2x - 1) + C(2x - 1) + D x^{2}. \] Expanding and equating coefficients gives \[ A = 2,\quad B = -3,\quad C = 4,\quad D = -1. \]

So \[ \boxed{\dfrac{4x^{3} - 9x^{2} + 11x - 4}{x^{2}(2x - 1)} = 2 - \frac{3}{x} + \frac{4}{x^{2}} - \frac{1}{2x - 1}}. \]

Question 6

(a) Factorise \(2x^{3} - 3x^{2} - 3x + 2\) completely.
(b) Hence express \[ \frac{x^{2} - 13x - 5}{2x^{3} - 3x^{2} - 3x + 2} \] in partial fractions.

Part (a)

Try simple roots: \(x = 2\) gives \(2(8) - 3(4) - 3(2) + 2 = 16 - 12 - 6 + 2 = 0\), so \(x = 2\) is a root and \((x - 2)\) is a factor.

Dividing \(2x^{3} - 3x^{2} - 3x + 2\) by \((x - 2)\) gives \((x + 1)(2x - 1)\), so \[ 2x^{3} - 3x^{2} - 3x + 2 = (x - 2)(x + 1)(2x - 1). \]

\[ \boxed{2x^{3} - 3x^{2} - 3x + 2 = (x - 2)(x + 1)(2x - 1)}. \]

Part (b)

Using the factorisation, write \[ \frac{x^{2} - 13x - 5}{(x - 2)(x + 1)(2x - 1)} = \frac{A}{x - 2} + \frac{B}{x + 1} + \frac{C}{2x - 1}. \] Solving for \(A,B,C\) gives \[ A = -3,\quad B = 1,\quad C = 5. \]

So \[ \boxed{\dfrac{x^{2} - 13x - 5}{2x^{3} - 3x^{2} - 3x + 2} = -\frac{3}{x - 2} + \frac{1}{x + 1} + \frac{5}{2x - 1}}. \]