Simple partial fractions | Year 12

01 / LEARN

The idea, step by step

Decompose an identity

For distinct linear denominator factors, write the proper rational expression as a sum of fractions over those factors. Clear denominators and compare coefficients or substitute convenient x values to find the constants. The equality must hold for every allowed x.

If the numerator degree is at least the denominator degree, divide first. Repeated factors need all necessary powers in the decomposition. Keep track of excluded denominator zeros throughout the calculation.

Integrate the simpler pieces

A term A/(x-a) integrates to $A ln ∣ x - a ∣$ . A repeated-factor power can use the power rule instead. Logarithmic absolute values keep the derivative valid on either side of a denominator zero.

For a definite integral, check that no pole lies inside the interval. Never apply endpoint subtraction across a point where the integrand is undefined. Differentiate the completed answer or recombine the partial fractions to check it.

A common trap

A decomposition is checked at just one x value.

Instead: It must be an identity. Clear denominators and compare all coefficients or test independent values sufficient to determine the constants.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

Solve $3 x + 5 = A (x + 2) + B (x + 1)$ to obtain A=2,B=1.
Integrate the two reciprocal terms.

2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ + C

Example 2

Worked example 2

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

Factor the denominator. The decomposition is $1 / [2 (x - 1)] + 1 / [2 (x + 1)]$ .
Integrate on an interval excluding -1 and 1.

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

Check your understanding

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

Show answer

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

The numerator after recombination is (x+1)-x=1.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

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

Read the worked solutions

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

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

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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.

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