Simple partial fractions | Year 12

01 / LEARN

Understand the idea.

Decompose an identity

Partial fractions replaces a proper rational expression with simpler fractions. For distinct linear factors, introduce one unknown numerator over each factor. Clear the denominators to obtain a polynomial identity, then compare coefficients or choose values that isolate the unknowns.

Check the rational expression before decomposing it. If the numerator degree is not smaller than the denominator degree, divide first. A repeated linear factor needs terms for every power up to its multiplicity, and all denominator zeros remain excluded.

Integrate the simpler pieces

Each simple reciprocal term integrates to a logarithm: $\int \frac{A}{x - a} d x = A ln ∣ x - a ∣ + C$ . The absolute value allows the derivative to work on either side of $a$ , although the antiderivative is used on an interval avoiding that point.

Recombine the fractions to check the identity before integrating. For definite integrals, inspect the whole interval for poles; ordinary endpoint subtraction cannot cross a denominator zero. Any improper integral requires a separate convergence argument.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Integrate $(3 x + 5) / [(x + 1) (x + 2)]$ .

Worked solution

1
Set up a decomposition.
$\frac{3 x + 5}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$
Both denominator factors are distinct and linear, and the rational expression is proper.
2
Determine the constants.
$\begin{aligned} 3 x + 5 & = A (x + 2) + B (x + 1) \\ x = - 1 : & 2 = A \\ x = - 2 : & - 1 = - B \end{aligned}$
Clearing denominators gives a polynomial identity, so convenient substituted values determine A and B.
3
Integrate the simpler fractions.
$\int (\frac{2}{x + 1} + \frac{1}{x + 2}) d x = 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + C$
Each reciprocal factor has a logarithmic antiderivative.
4
Check the identity and domain.
$2 (x + 2) + (x + 1) = 3 x + 5, x \neq - 1, - 2$
The numerators recombine correctly; use the antiderivative on an interval avoiding both poles.

Answer

$2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + C$ .

EXAMPLE 02

Worked example 2

Integrate $x / (x^{2} - 1)$ .

Worked solution

1
Factor and introduce constants.
$\frac{x}{x^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}, x \neq \pm 1$
The denominator factors as a difference of squares.
2
Solve the identity.
$\begin{array}{c} x = A (x + 1) + B (x - 1) \\ A + B = 1, A - B = 0 \\ A = B = \frac{1}{2} \end{array}$
Matching the linear and constant coefficients fixes both constants.
3
Integrate and retain the restrictions.
$\int \frac{x}{x^{2} - 1} d x = \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + C$
This formula differentiates to the decomposed fractions on any interval excluding minus one and one.

Answer

$\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + C$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Decompose 1/[x(x+1)].

Show answer & reasoning

$1 / x - 1 / (x + 1)$ .

Recombining $\frac{1}{x} - \frac{1}{x + 1}$ gives numerator $(x + 1) - x = 1$ over $x (x + 1)$ . Hence the decomposition is valid for $x \neq 0, - 1$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Integrate $\frac{1}{x (x + 1)}$ .
2 marks
Integrate $\frac{2}{x^{2} - 4}$ .
2 marks
Integrate $\frac{5 x + 1}{(x - 1) (x + 1)}$ .
2 marks
Integrate $\frac{2 x + 3}{{(x + 1)}^{2}}$ .
2 marks
Integrate $\frac{x^{2} + 1}{x^{2} - 1}$ .
2 marks
Evaluate $\int_{1}^{2} \frac{1}{x (x + 1)} d x$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Integrate $\frac{1}{x (x + 1)}$ .
2 marks
Use the difference of reciprocal terms.
Integrate each and exclude $x = 0, - 1$ .
$ln ∣ x ∣ - ln ∣ x + 1 ∣ + C$ .
Integrate $\frac{2}{x^{2} - 4}$ .
2 marks
Decompose as $\frac{1}{2 (x - 2)} - \frac{1}{2 (x + 2)}$ .
Integrate away from the poles.
$\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x + 2 ∣ + C$ .
Integrate $\frac{5 x + 1}{(x - 1) (x + 1)}$ .
2 marks
Use coefficients $A = 3, B = 2$ .
Their cleared numerator is $3 (x + 1) + 2 (x - 1)$ .
$3 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ + C$ .
Integrate $\frac{2 x + 3}{{(x + 1)}^{2}}$ .
2 marks
Write $\frac{2}{x + 1} + \frac{1}{{(x + 1)}^{2}}$ .
Integrate by logarithm and power rules.
$2 ln ∣ x + 1 ∣ - \frac{1}{x + 1} + C$ .
Integrate $\frac{x^{2} + 1}{x^{2} - 1}$ .
2 marks
Divide to obtain $1 + \frac{2}{x^{2} - 1}$ .
Decompose the remaining fraction.
$x + ln ∣ x - 1 ∣ - ln ∣ x + 1 ∣ + C$ .
Evaluate $\int_{1}^{2} \frac{1}{x (x + 1)} d x$ .
2 marks
Use $ln x - ln (x + 1)$ on this positive interval.
Endpoint subtraction gives $ln (\frac{2}{3}) - ln (\frac{1}{2})$ .
$ln (\frac{4}{3})$ .

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

In this lesson

Understand the idea.

Decompose an identity

Integrate the simpler pieces

See the method in action.

Worked example 1

Set up a decomposition.

Determine the constants.

Integrate the simpler fractions.

Check the identity and domain.

Worked example 2

Factor and introduce constants.

Solve the identity.

Integrate and retain the restrictions.

Try it yourself.

Check your understanding

Keep practising