# Partial Fractions

## Partial Fractions Revision

**Partial Fractions**

Algebraic fractions that have a denominator with **more than one linear factor** can be split, *or decomposed*, into **partial fractions**. This can be particularly useful when integrating complex expressions or with binomial expansions.

Questions involving partial fractions will only ever involve a **numerator** that is a **constant**Â or a **linear** term and the **denominator** will only ever include up to **three** **terms** with the most complex being **squared** **linear** **terms**.

**Types of Partial Fraction**

There are **3** types of partial fraction that you will see in A level Maths:

**1.** Considering a fraction where the denominator has **two** **linear** **terms**, we can split up the fraction into two separate fractions,

Â \dfrac{5}{\textcolor{limegreen}{(x+1)} \textcolor{blue}{(x+2)}} can be split up into \dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}}

**2.** Considering a fraction where the denominator has **three** **linear** **terms**, we can split up the fraction into three separate fractions,

\dfrac{3x+4}{\textcolor{limegreen}{(x+1)}\textcolor{blue}{(x+2)}\textcolor{orange}{(x-1)}} can be split up into \dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}} +\dfrac{C}{\textcolor{orange}{(x-1)}}

**3.** Considering a fraction where the denominator has **one** **linear** **term** and **a repeated term**, we can split up the fraction into three separate fractions,

\dfrac{2x+1}{{\textcolor{limegreen}{(x+1)}}^2 \textcolor{blue}{(x+2)}} can be split up into \dfrac{A}{{\textcolor{limegreen}{(x+1)}}^2}+\dfrac{B}{\textcolor{limegreen}{(x+1)}} +\dfrac{C}{\textcolor{blue}{(x+2)}}

**Expressing Fractions as Partial Fractions**

To find the coefficients in the numerators in the partial fractions, A, B and C, there are **2** different methods you can use: **Substitution** or **Equating Coefficients**.

**Example:** Express \dfrac{5x+1}{(x-1)(x+1)(x+2)} as partial fractions.

**Step 1:** Write the fraction as partial fractions with unknown constants, and put it over a common denominator:

\begin{aligned} \dfrac{5x+1}{(x-1)(x+1)(x+2)} &\equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x+1)} + \dfrac{C}{(x+2)} \\[1.2em] &\equiv \dfrac{A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)}{(x-1)(x+1)(x+2)} \end{aligned}

**Step 2:** Cancel down the denominators:

5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)

**Step 3:** Use the **Substitution** or **Equating Coefficients** to find the values of the unknown constants, A, B and C:

**Substitution:**

Substitute values of x inside the brackets to make them 0, so that you are left with only one of A, B and C:

Substitute in x = 1:

\begin{aligned} 5(1) + 1 &= A(1+1)(1+2) \\ 6 &= 6A \\ \textcolor{red}{A} &\textcolor{red}{=} \textcolor{red}{1} \end{aligned}

Substitute in x = -1:

\begin{aligned} 5(-1) + 1 &= B(-1-1)(-1+2) \\ -4 &= -2B \\ \textcolor{blue}{B} &\textcolor{blue}{=} \textcolor{blue}{2} \end{aligned}

Substitute in x = -2:

\begin{aligned} 5(-2) + 1 &= C(-2-1)(-2+1) \\ -9 &= 3C \\ \textcolor{limegreen}{C} &\textcolor{limegreen}{=} \textcolor{limegreen}{-3} \end{aligned}

**Equating Coefficients:**

Equate coefficients and then solve them simultaneously to find the values of the unknown constants, A, B and C:

5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)

\equiv A(x^2 + 3x + 2) + B(x^2 + x - 2) + C(x^2 - 1)

\equiv (A + B + C)x^2 + (3A + B)x + (2A - 2B - C)

Equating x^2 coefficients gives:

0 = A + B + C

Equating x coefficients gives:

5 = 3A + B

Equating constant terms gives:

1 = 2A - 2B - C

Then, solve these simultaneously, which gives \textcolor{red}{A = 1}, \textcolor{blue}{B = 2} and \textcolor{limegreen}{C = -3}

**Step 4:** Replace the constant terms, A, B and C in the original identity:

\dfrac{5x+1}{(x-1)(x+1)(x+2)} \equiv \dfrac{\textcolor{red}{1}}{(x-1)} + \dfrac{\textcolor{blue}{2}}{(x+1)} \textcolor{limegreen}{-} \dfrac{\textcolor{limegreen}{3}}{(x+2)}

**Note:** The unknown constants may not always be this nice – you may get questions with constant terms that are fractions.

**Note:**

- You may need to combine the methods of Substitution and Equating Coefficients for some examples, such as fractions with repeated terms.
- You may be given a fraction that is not split up into brackets in the denominator, so you will need to factorise it. Also, look out for the difference of two squares.

**Example: Repeated Terms**

Express \dfrac{2x-3}{x(x-2)^2} as partial fractions.

**[3 marks]**

Write the fraction as partial fractions with unknown constants, and put it over a common denominator:

\begin{aligned} \dfrac{2x-3}{x(x-2)^2} &\equiv \dfrac{A}{(x-2)^2} + \dfrac{B}{x-2} + \dfrac{C}{x} \\[1.1em] &\equiv \dfrac{Ax + Bx(x-2) + C(x-2)^2}{x(x-2)^2} \end{aligned}

Then, cancel the denominators:

2x-3 \equiv Ax + Bx(x-2) + C(x-2)^2

Substitute in x=0:

\begin{aligned} -3 &= C(-2)^2 \\ C &= - \dfrac{3}{4} \end{aligned}

Substitute in x=2:

\begin{aligned} 2(2) - 3 &= 2A \\ A &= \dfrac{1}{2} \end{aligned}

Now, equate coefficients of x^2:

\begin{aligned} 0 &= B + C \\ B &= \dfrac{3}{4} \end{aligned}

Finally, replace the values of A, B and C into the original identity:

\dfrac{2x-3}{x(x-2)^2} \equiv \dfrac{1}{2(x-2)^2} + \dfrac{3}{4(x-2)} - \dfrac{3}{4x}

## Partial Fractions Example Questions

**Question 1:** Express \dfrac{8}{9x^2 - 16} as partial fractions.

**[3 marks]**

\dfrac{8}{9x^2 - 16} = \dfrac{8}{(3x+4)(3x-4)}

\begin{aligned} \dfrac{8}{(3x+4)(3x-4)} &\equiv \dfrac{A}{(3x+4)} + \dfrac{B}{(3x-4)} \\[1.1em] &\equiv \dfrac{A(3x-4) + B(3x+4)}{(3x+4)(3x-4)} \end{aligned}

8 \equiv A(3x-4) + B(3x+4)

Substitute in x=\dfrac{4}{3}:

\begin{aligned} 8 &= 8B \\ B &= 1 \end{aligned}

Substitute in x=- \dfrac{4}{3}:

\begin{aligned} 8 &= -8A \\ A &= -1 \end{aligned}

Then, replace the values of A and B into the identity:

\dfrac{8}{9x^2 - 16} \equiv - \dfrac{1}{(3x+4)} + \dfrac{1}{(3x-4)}

**Question 2:** Express \dfrac{x}{(x-1)(x-2)(x-3)} as partial fractions.

**[3 marks]**

\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x-2)} + \dfrac{C}{(x-3)}

\equiv \dfrac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}

x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)

Substitute in x = 1:

\begin{aligned} 1 &= A(1-2)(1-3) \\ A &= \dfrac{1}{2} \end{aligned}

Substitute in x = 2:

\begin{aligned} 2 &= B(2-1)(2-3) \\ B &= -2 \end{aligned}

Substitute in x = 3:

\begin{aligned} 3 &= C(3-1)(3-2) \\ C &= \dfrac{3}{2} \end{aligned}

Then, replace the values of A, B and C into the identity:

\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{1}{2(x-1)} - \dfrac{2}{(x-2)} + \dfrac{3}{2(x-3)}

**Question 3:** Express \dfrac{10}{x^2(x+1)} as partial fractions.

**[3 marks]**

10 \equiv A(x+1) + Bx(x+1) + Cx^2

Substitute in x = 0:

\begin{aligned} 10 &= A(0+1) \\ A &= 10 \end{aligned}

Substitute in x = -1:

\begin{aligned} 10 &= C(-1)^2 \\ C &= 10 \end{aligned}

Equate coefficients of x^2:

\begin{aligned} 0 &= B + C \\ B &= -10 \end{aligned}

Then, replace the values of A, B and C into the identity:

\dfrac{10}{x^2(x+1)} \equiv \dfrac{10}{x^2} - \dfrac{10}{x} + \dfrac{10}{(x+1)}

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