1.02y - Partial fractions with linear factors | OCR A-Level Mathematics A

Split a complicated fraction into simple pieces, including repeated linear factors, and check the result by recombining. Then use the split with supplied expansion and integration rules to make longer problems manageable.

Splitting two distinct linear factors

Can making one fraction into several actually make a problem easier? Compare $1/((1-x)(1-2x))$ with $-1/(1-x)+2/(1-2x)$ . At $x=0$ , both give $1$ ; at $x=1/4$ , both give $8/3$ . The second form will make its power series almost immediate.

You already know

Adding fractions uses a common denominator. A polynomial identity holds at every real input; a fraction equation still excludes every zero of its denominator.

We are going to reverse that common-denominator calculation.

Partial fractions

A partial-fraction decomposition expresses a rational expression as a sum of simpler fractions with the same values on its domain. Distinct linear denominator factors each receive one constant numerator. For example, a denominator $(x-1)(x+2)$ uses $A/(x-1)+B/(x+2)$ .

The constants are unknown until the numerator identity determines them.

Worked example 1: two monic factors

Question. Express $(5x+1)/((x-1)(x+2))$ in partial fractions.

Plan. Clear denominators and substitute the zeros of the factors into the resulting polynomial identity.

\begin{aligned} 5x+1&\equiv A(x+2)+B(x-1) && \text{clear denominators}\\ x=1:\quad6&=3A\quad\Rightarrow\quad A=2 && \text{remove }B\\ x=-2:\quad-9&=-3B\quad\Rightarrow\quad B=3 && \text{remove }A \end{aligned}

Why this step? Although the fractions are undefined at $1,-2$ , the cleared polynomial identity holds there. We substitute into the identity, not into an undefined fraction.

Answer. $2/(x-1)+3/(x+2)$ , for $x\ne1,-2$ .

Check. The recombined numerator is $2(x+2)+3(x-1)=5x+1$ .

Change one feature: one factor has leading coefficient $2$ . Its zero and its multiplier must both be respected.

Worked example 2: a non-monic factor

Question. Express $(7x+4)/((2x-1)(x+2))$ in partial fractions.

Plan. Keep the factors as written and use $x=1/2,-2$ .

\begin{aligned} 7x+4&\equiv A(x+2)+B(2x-1) && \text{clear denominators}\\ x=\tfrac12:\quad\tfrac{15}2&=\tfrac52 A\quad\Rightarrow\quad A=3 && \text{first constant}\\ x=-2:\quad-10&=-5B\quad\Rightarrow\quad B=2 && \text{second constant} \end{aligned}

Why this step? Setting $x=1$ would not make $2x-1$ zero; $1/2$ does.

Answer. $3/(2x-1)+2/(x+2)$ , for $x\ne1/2,-2$ .

Check. $3(x+2)+2(2x-1)=7x+4$ .

You can instead compare coefficients. In the last identity, $A+2B=7$ and $2A-B=4$ . Both routes give the same constants. The decomposition method is not in the formula sheet: learn it.

Watch out

Wrong: $1/(2x-1)=1/(x-1/2)$ . Right: $1/(2x-1)=(1/2)/(x-1/2)$ . If you rescale a denominator, rescale its numerator too.

Keep the non-monic factor but change the numerator.

Handling three distinct factors

Three distinct linear factors need three simple fractions. Clearing the denominator gives a quadratic identity even when the starting numerator is only constant or linear.

Worked example 3: a constant numerator

Question. Decompose $6/((x-1)(x+1)(x+2))$ .

Plan. Put constants $A,B,C$ over the three factors and isolate each in turn.

\begin{aligned} 6&\equiv A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1) && \text{clear factors}\\ x=1:\quad6&=6A\quad\Rightarrow\quad A=1 && \text{isolate }A\\ x=-1:\quad6&=-2B\quad\Rightarrow\quad B=-3 && \text{isolate }B\\ x=-2:\quad6&=3C\quad\Rightarrow\quad C=2 && \text{isolate }C \end{aligned}

Why this step? At each chosen input two products vanish, leaving just one constant to find.

Answer. $1/(x-1)-3/(x+1)+2/(x+2)$ , with $x\ne1,-1,-2$ .

Check. $(x^2+3x+2)-3(x^2+x-2)+2(x^2-1)=6$ . Both higher-power coefficients cancel.

Change one feature: make the numerator linear. The denominator and setup stay the same.

Worked example 4: a linear numerator

Question. Decompose $(x+5)/((x-1)(x+1)(x+2))$ .

Plan. Use the same identity with $x+5$ on its left.

\begin{aligned} x=1:\quad6&=6A\quad\Rightarrow\quad A=1 && \text{first constant}\\ x=-1:\quad4&=-2B\quad\Rightarrow\quad B=-2 && \text{second constant}\\ x=-2:\quad3&=3C\quad\Rightarrow\quad C=1 && \text{third constant} \end{aligned}

Why this step? The new numerator changes the values substituted on the left; reusing the previous constants would give the wrong identity.

Answer. $1/(x-1)-2/(x+1)+1/(x+2)$ , for $x\ne1,-1,-2$ .

Check. $(x^2+3x+2)-2(x^2+x-2)+(x^2-1)=x+5$ .

Our denominator is no more complicated than three distinct linear factors or one linear factor times a squared linear factor. Numerators here are constant or linear, and the split uses no more than three terms. This setup is not in the formula sheet: learn it. An irreducible quadratic such as $x^2+1$ needs a different setup, beyond this lesson's scope.

Problem-solving

Check the recombined numerator, including coefficients that should be zero. With a constant starting numerator, all $x^2$ and $x$ terms must cancel.

Keep a linear numerator and change its constant.

Including every power of a repeated factor

A squared factor needs two available terms: one over its first power and one over its square. Without both, the numerator usually cannot be reconstructed.

A squared linear factor

For distinct factor zeros, the complete form is

\frac{px+q}{(ax+b)(cx+d)^2} =\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2},

where $a,c\ne0$ , and all denominator zeros remain excluded. Some solved coefficients may be zero.

Not in the formula sheet: learn it.

Use substitution for the constants it isolates, then compare a coefficient for the remaining one.

Worked example 5: a squared monic factor

Question. Decompose $(3x+1)/((x-1)(x+1)^2)$ .

Plan. Include both powers of $x+1$ , clear the denominator and use the $x^2$ coefficient last.

\begin{aligned} 3x+1&\equiv A(x+1)^2+B(x-1)(x+1)+C(x-1) && \text{complete identity}\\ x=1:\quad4&=4A\quad\Rightarrow\quad A=1 && \text{isolate }A\\ x=-1:\quad-2&=-2C\quad\Rightarrow\quad C=1 && \text{isolate }C\\ 0&=A+B\quad\Rightarrow\quad B=-1 && \text{compare }x^2 \end{aligned}

Why this step? Substitution at the repeated zero removes both terms involving $A,B$ , so a different equation is needed for $B$ .

Answer. $1/(x-1)-1/(x+1)+1/(x+1)^2$ , for $x\ne1,-1$ .

Check. $(x+1)^2-(x-1)(x+1)+(x-1)=3x+1$ .

Change one feature: replace the repeated monic factor by $2x+1$ . Its zero is now $-1/2$ .

Worked example 6: a squared non-monic factor

Question. Decompose $(7x+2)/((x-1)(2x+1)^2)$ .

Plan. Keep both powers of $2x+1$ and compare the quadratic coefficient.

\begin{aligned} 7x+2&\equiv A(2x+1)^2+B(x-1)(2x+1)+C(x-1) && \text{clear denominator}\\ x=1:\quad9&=9A\quad\Rightarrow\quad A=1 && \text{first constant}\\ x=-\tfrac12:\quad-\tfrac32&=-\tfrac32 C\quad\Rightarrow\quad C=1 && \text{third constant}\\ 0&=4A+2B\quad\Rightarrow\quad B=-2 && \text{compare }x^2 \end{aligned}

Why this step? Squaring $2x+1$ makes its leading coefficient $4$ , which matters in the coefficient equation.

Answer. $1/(x-1)-2/(2x+1)+1/(2x+1)^2$ , with $x\ne1,-1/2$ .

Check. $(2x+1)^2-2(x-1)(2x+1)+(x-1)=7x+2$ .

Leaving out a possible term can accidentally work in a special case, but it is not a reliable general method. Set up every allowed power before finding the constants.

Watch out

Wrong: use only $A/(x-1)+C/(x+1)^2$ for a general denominator $(x-1)(x+1)^2$ . Right: include $B/(x+1)$ too. Its coefficient may be non-zero.

Keep the non-monic repeated factor and change the numerator.

Using the split in a power series

Partial fractions let you expand one simple factor at a time. We supply the needed rules here; the series lesson develops them further.

General binomial expansion

For real $n$ and $|u|<1$ ,

(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\cdots.

For non-negative integer $n$ , the terminating expansion is valid for every real $u$ .

In the formula sheet.

Set $n=-1,-2$ and replace $u$ by $-u$ . The useful cases, not printed separately on the sheet, are

\frac1{1-u}=1+u+u^2+\cdots,\qquad \frac1{(1-u)^2}=1+2u+3u^2+\cdots,\qquad |u|<1.

A power series is an infinite sum of powers; writing terms through $x^2$ truncates it and is not an exact polynomial identity.

Worked example 7: two simple factors

Question. Find the expansion through $x^2$ of $1/((1-x)(1-2x))$ , using its split $-1/(1-x)+2/(1-2x)$ . State validity.

Plan. Expand each fraction separately and intersect their validity conditions.

\begin{aligned} -\frac1{1-x}+\frac2{1-2x} &=-(1+x+x^2+\cdots)+2(1+2x+4x^2+\cdots) && \text{use }u=x,2x\\ &=1+3x+7x^2+\cdots && \text{collect coefficients} \end{aligned}

Why this step? Both expansions must work together: $|x|<1$ and $|2x|<1$ .

Answer. $1+3x+7x^2$ through $x^2$ , valid for $|x|<1/2$ .

Check. Multiplying the displayed terms by $1-3x+2x^2$ gives constant $1$ and zero coefficients of $x,x^2$ .

Change one feature: square the factor $1-2x$ . Its coefficients change, but its zero and the strict validity bound do not.

Worked example 8: a repeated factor in a series

Question. Expand $1/((1-x)(1-2x)^2)$ through $x^2$ , using

\frac1{(1-x)(1-2x)^2}=\frac1{1-x}-\frac2{1-2x}+\frac2{(1-2x)^2}.

Plan. Use the squared-factor rule for the last term and collect powers.

\begin{aligned} & (1+x+x^2+\cdots)-2(1+2x+4x^2+\cdots) && \text{simple factors}\\ &\quad+2(1+4x+12x^2+\cdots) && \text{squared factor}\\ &=1+5x+17x^2+\cdots && \text{collect powers} \end{aligned}

Why this step? The squared rule contributes $3(2x)^2=12x^2$ , not $4x^2$ .

Answer. $1+5x+17x^2$ through $x^2$ , for $|x|<1/2$ .

Check. The split recombines because $(1-2x)^2-2(1-x)(1-2x)+2(1-x)=1$ . Its denominator is $1-5x+8x^2-4x^3$ ; the product with the displayed series has no $x$ or $x^2$ term.

The opening fractions really were easier to use separately. The striking coefficients $1,3,7$ are $2^1-1,2^2-1,2^3-1$ ; in fact its coefficient of $x^n$ is $2^{n+1}-1$ , directly from the two geometric expansions.

Watch out

Wrong: use the rational expression's whole domain as the series interval. Right: intersect the expansion conditions. A fraction may exist at an input where this power series does not converge.

Change the repeated factor's scale, using the supplied split.

Using the split in an integration step

An antiderivative differentiates back to the integrand. Splitting a fraction lets you apply familiar rules to each piece; we supply those rules here so the focus stays on using the decomposition.

Integration rules

For a differentiable function $\mathrm f$ , on an interval where $\mathrm f(x)\ne0$ ,

\int\frac{\mathrm f'(x)}{\mathrm f(x)}\,\mathrm dx=\ln|\mathrm f(x)|+C,

and, for $n\ne-1$ ,

\int\mathrm f'(x)(\mathrm f(x))^n\,\mathrm dx=\frac{(\mathrm f(x))^{n+1}}{n+1}+C.

Use real-valued powers on their domain.

In the formula sheet.

For $a\ne0$ , these give the following useful cases, not printed separately on the sheet. We supply them for this application:

\int\frac1{ax+b}\,\mathrm dx=\frac1a\ln|ax+b|+C, \qquad \int\frac1{(ax+b)^2}\,\mathrm dx=-\frac1{a(ax+b)}+C.

Work on an interval not crossing $ax+b=0$ .

The factor $1/a$ compensates for the derivative of $ax+b$ . One arbitrary constant covers the sum on each allowed interval.

Worked example 9: integrate simple factors

Question. Integrate $3x/((x-1)(2x+1))$ , given its split $1/(x-1)+1/(2x+1)$ .

Plan. Apply the logarithm rule to each term, including its denominator coefficient.

\begin{aligned} \int\frac{3x}{(x-1)(2x+1)}\,\mathrm dx &=\int\left(\frac1{x-1}+\frac1{2x+1}\right)\mathrm dx && \text{use the split}\\ &=\ln|x-1|+\tfrac12\ln|2x+1|+C && \text{integrate terms} \end{aligned}

Why this step? Differentiating $\ln|2x+1|$ introduces $2$ , so its coefficient must be $1/2$ .

Answer. $\ln|x-1|+\frac12\ln|2x+1|+C$ , on an interval avoiding $1,-1/2$ .

Check. Differentiating gives $1/(x-1)+1/(2x+1)$ ; recombining gives numerator $(2x+1)+(x-1)=3x$ .

Change one feature: the second factor is squared. It now needs a reciprocal antiderivative as well as a logarithm.

Worked example 10: integrate a repeated factor

Question. Integrate $(7x+2)/((x-1)(2x+1)^2)$ , using the earlier split.

Plan. Integrate $1/(x-1)-2/(2x+1)+1/(2x+1)^2$ term by term.

\begin{aligned} \int\left(\frac1{x-1}-\frac2{2x+1}+\frac1{(2x+1)^2}\right)\mathrm dx &=\ln|x-1|-\ln|2x+1|\\ &\quad-\frac1{2(2x+1)}+C && \text{apply both rules} \end{aligned}

Why this step? A squared denominator uses the reciprocal rule, not the logarithm rule.

Answer. $\ln|x-1|-\ln|2x+1|-1/(2(2x+1))+C$ , on an interval avoiding $1,-1/2$ .

Check. The derivative of the last term is $1/(2x+1)^2$ . The two logarithms give the other two fractions, already verified by recombination.

For comparison, the monic split from Worked example 1 gives

\int\frac{5x+1}{(x-1)(x+2)}\,\mathrm dx=2\ln|x-1|+3\ln|x+2|+C.

The absolute values allow either sign of each non-zero argument; an integral is taken on an interval without a denominator zero. Constants can differ on disconnected intervals.

Watch out

Wrong: integrate every denominator using a logarithm. Right: a first power produces a logarithm; a squared linear factor produces a reciprocal. Differentiate your answer to catch missing factors and signs.

Keep the repeated-factor split, with a new coefficient in the last term.

Quick Recap

Distinct linear factors each need one constant-numerator fraction.
A squared factor needs both its first and second powers; include all possible terms before solving.
Clear denominators to obtain a polynomial identity, then substitute factor zeros or compare coefficients and recombine to check.
A split makes expansions and integrals simpler; keep the correct series interval and integration domain.

Yes, several fractions can be easier than one: the opening split gives $1+3x+7x^2+\cdots$ immediately, for $|x|<1/2$ .

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1.02y - Partial fractions with linear factors

Splitting two distinct linear factors

Handling three distinct factors

Including every power of a repeated factor

Using the split in a power series

Using the split in an integration step

Quick Recap

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