Pure 8.6 - Integration using partial fractions | Edexcel A-Level Mathematics

One rational expression can hide several easy integrals. You will split linear denominator factors into partial fractions, integrate logarithmic and repeated-factor terms, and evaluate exact definite integrals while keeping coefficients, signs and domains under control.

Why partial fractions help integration

How can splitting one fraction make integration easier? Consider

\frac{7x+1}{(x-2)(x+3)}=\frac3{x-2}+\frac4{x+3}.

The left side is unfamiliar as an integral; the right side is two logarithmic derivatives. Recombining the right side gives $3(x+3)+4(x-2)=7x+1$ , so no approximation has occurred.

You already know

Partial fractions rewrite a rational expression as a sum. Use $A/(x-a)+B/(x-b)$ for distinct factors and include both $B/(x-b)$ and $D/(x-b)^2$ when a factor is squared. Work away from denominator zeros.

Integration is linear: integrate each summand, then add. Higher denominator powers behave differently from the first power.

Integrating linear-factor terms

For constants $A,a,b$ , with $a\ne0$ , on an interval avoiding $ax+b=0$ ,

\int\frac{A}{ax+b}\,\mathrm{d}x=\frac Aa\ln|ax+b|+C.

For an integer $n>1$ ,

\int\frac{A}{(ax+b)^n}\,\mathrm{d}x=\frac{A}{a(1-n)}(ax+b)^{1-n}+C.

The second result is the power rule with a chain factor. Not in the formula booklet: learn it.

For example, $3/(x-2)^2=3(x-2)^{-2}$ integrates to $-3/(x-2)+C$ , whereas $3/(x-2)$ integrates to $3\ln|x-2|+C$ . The exponent decides the rule.

Interval of integration

Here an interval of integration must avoid every zero of the original denominator. An antiderivative formula can apply on several separate intervals, with its own arbitrary constant on each. A formula written on both sides of a pole does not make integration across that pole legitimate.

There is a useful surprise: terms that each resemble $1/x$ can cancel their slow decay. For $x>0$ ,

\frac1x-\frac1{x+5}=\frac5{x(x+5)}.

Multiplying by $x^2$ gives $5x/(x+5)$ , which tends to $5$ . The difference therefore decays like $5/x^2$ , even though the individual terms decay like $1/x$ .

Key idea

Decompose first, then choose the integral rule separately for each denominator power.

Diagnose the tempting wrong rule.

Integrate with distinct linear factors

For distinct linear factors, the constants become the logarithm coefficients when the factor gradients are $1$ . Find the decomposition before integrating.

Worked example 1: two distinct factors

Question. Find $\int(7x+1)/[(x-2)(x+3)]\,\mathrm{d}x$ .

Plan. Write two partial fractions, determine their constants at the factor zeros, then integrate.

\begin{aligned} 7x+1&\equiv A(x+3)+B(x-2) && \text{clear denominators}\\ x=2&:\ 15=5A\Rightarrow A=3 && \text{remove the }B\text{ term}\\ x=-3&:\ -20=-5B\Rightarrow B=4 && \text{remove the }A\text{ term}\\ \int\frac{7x+1}{(x-2)(x+3)}\,\mathrm{d}x&=\int\left(\frac3{x-2}+\frac4{x+3}\right)\,\mathrm{d}x && \text{split the integral} \end{aligned}

Why those two values? Each makes one factor vanish and isolates one constant in the polynomial identity.

Answer. $3\ln|x-2|+4\ln|x+3|+C$ , on an interval avoiding $-3,2$ .

Check. Differentiating and recombining gives numerator $3(x+3)+4(x-2)=7x+1$ .

Change only the numerator. The denominator factors and the integration rules stay the same.

Worked example 2: a changed numerator

Question. Find $\int(9x+7)/[(x-2)(x+3)]\,\mathrm{d}x$ .

Plan. Use the same partial-fraction form and the same two substitutions.

\begin{aligned} 9x+7&\equiv A(x+3)+B(x-2) && \text{clear denominators}\\ x=2&:\ 25=5A\Rightarrow A=5 && \text{first constant}\\ x=-3&:\ -20=-5B\Rightarrow B=4 && \text{second constant}\\ \int\frac{9x+7}{(x-2)(x+3)}\,\mathrm{d}x&=5\ln|x-2|+4\ln|x+3|+C && \text{integrate separately} \end{aligned}

Why keep both terms? The numerator has not cancelled either denominator factor.

Answer. $5\ln|x-2|+4\ln|x+3|+C$ .

Check. $5(x+3)+4(x-2)=9x+7$ .

Use one arbitrary constant for the whole sum: two additive constants would simply combine.

Exam tip

Show the identity used to find the constants and then the integrated expression. A correct decomposition alone has not completed an integration question.

Change the numerator again while retaining the two factors.

Keep the chain factor and sign

The coefficient above a linear factor is not automatically the coefficient of its logarithm. Compare $\int6/(3x+2)\,\mathrm{d}x=2\ln|3x+2|+C$ with $\int6/(2-3x)\,\mathrm{d}x=-2\ln|2-3x|+C$ . The inner derivatives are $3$ and $-3$ .

Worked example 3: a non-unit factor gradient

Question. Find $\int(13x+4)/[(x+2)(3x-1)]\,\mathrm{d}x$ .

Plan. Decompose, then divide each logarithm coefficient by its factor gradient.

\begin{aligned} 13x+4&\equiv A(3x-1)+B(x+2) && \text{clear denominators}\\ x=-2&:\ -22=-7A\Rightarrow A=22/7 && \text{first constant}\\ x=1/3&:\ 25/3=7B/3\Rightarrow B=25/7 && \text{second constant}\\ \int\frac{13x+4}{(x+2)(3x-1)}\,\mathrm{d}x&=\frac{22}{7}\ln|x+2|+\frac{25}{21}\ln|3x-1|+C && \text{divide by }1\text{ and }3 \end{aligned}

Why $25/21$ ? Differentiation of $\ln|3x-1|$ multiplies its coefficient by $3$ .

Answer. $\frac{22}{7}\ln|x+2|+\frac{25}{21}\ln|3x-1|+C$ .

Check. Its derivative is $22/[7(x+2)]+25/[7(3x-1)]$ ; the combined numerator is $13x+4$ .

Change only the orientation of the second factor from $3x-1$ to $1-3x$ . The original rational function changes sign, which is a helpful final check.

Worked example 4: a negative factor gradient

Question. Find $\int(13x+4)/[(x+2)(1-3x)]\,\mathrm{d}x$ .

Plan. Find the constants for the reversed factor, then use its derivative $-3$ .

\begin{aligned} 13x+4&\equiv A(1-3x)+B(x+2) && \text{clear denominators}\\ x=-2&:\ -22=7A\Rightarrow A=-22/7 && \text{first constant}\\ x=1/3&:\ 25/3=7B/3\Rightarrow B=25/7 && \text{second constant}\\ \int\frac{13x+4}{(x+2)(1-3x)}\,\mathrm{d}x&=-\frac{22}{7}\ln|x+2|-\frac{25}{21}\ln|1-3x|+C && \text{chain factor }-1/3 \end{aligned}

Why a negative second logarithm coefficient? The denominator's derivative is negative, even though $B$ is positive.

Answer. $-\frac{22}{7}\ln|x+2|-\frac{25}{21}\ln|1-3x|+C$ .

Check. Since $|1-3x|=|3x-1|$ , this antiderivative is the negative of Worked example 3, as required.

A compact version of the same sign decision is $\int4/(5-2x)\,\mathrm{d}x=-2\ln|5-2x|+C$ .

Watch out

Wrong: integrating $B/(3x-1)$ as $B\ln|3x-1|$ . Right: $(B/3)\ln|3x-1|$ . Differentiate the proposed answer to expose the extra factor immediately.

Double only the numerator of the second example.

Integrate repeated linear factors

A repeated factor produces both a logarithm term and a negative-power term. Keep them separate during integration.

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Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 8.6 - Integration using partial fractions

Why partial fractions help integration

Integrate with distinct linear factors

Keep the chain factor and sign

Integrate repeated linear factors

Continue this lesson.

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