Mathematics 9709 · Pure Mathematics 3

Integrating rational functions using partial fractions

This is the most common type in the chapter: partial fractions turn up in an integration question in about 30 of 42 recent papers.
The wording is almost always “express in partial fractions, hence find the exact value in the form $a+b\ln c$ .”
Get this right and you pick up the biggest single block of marks in the chapter.

Split first, then integrate one piece at a time

The algebra of splitting the fraction is Chapter 7 work, so take it as known. (Distinct linear → $\tfrac{A}{x-p}+\tfrac{B}{x-q}$ ; repeated linear → $\tfrac{A}{x-p}+\tfrac{B}{(x-p)^2}$ ; quadratic $cx^2+d$ → $\tfrac{Bx+C}{x^2+d}$ ; improper → divide first.)
This topic is about integrating each piece once it is split.

How to spot it: a factored bottom with a fraction on top → split into partial fractions (Ch 7), then integrate each piece one at a time.

Partial-fraction piece	Integrates to
$\dfrac{A}{x-p}$	$A\ln\|x-p\|$
$\dfrac{B}{(x-p)^2}$	$-\dfrac{B}{x-p}$ (power rule — NOT a log)
$\dfrac{Bx+C}{x^2+d}$	split into $\tfrac{B}{2}\ln(x^2+d)$ + $\tfrac{C}{\sqrt d}\tan^{-1}\!\frac{x}{\sqrt d}$

The main trap: a repeated factor is a reciprocal, not a log

Every linear factor on the bottom integrates to a $\ln$ , except the squared one.
A $\dfrac{B}{(x-p)^2}$ term is $B(x-p)^{-2}$ , and the power rule turns it into $-B(x-p)^{-1}=-\dfrac{B}{x-p}$ .
It is not a second log.

Common mistake

Integrating

\dfrac{B}{(x-p)^2}

B\ln|x-p|

is the most common slip the markscheme flags in this chapter. A repeated factor

(x-p)^2

on the bottom integrates to

-\dfrac{B}{x-p}

by the power rule.

Worked example

9709/31 Jun 2023: $f(x)=\dfrac{3-3x^2}{(2x+1)(x+2)^2}$

(a) Decompose (Ch 7 work) into $\dfrac{A}{2x+1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$ .
(b) Integrate term by term: the $\dfrac{A}{2x+1}$ → $\tfrac{A}{2}\ln|2x+1|$ , the $\dfrac{B}{x+2}$ → $B\ln|x+2|$ , and the repeated $\dfrac{C}{(x+2)^2}$ → $-\dfrac{C}{x+2}$ (reciprocal, not a log).
Put in the limits and collapse the logs into the required $a+b\ln c$ form using log laws.

(9709/31 Jun 2023 Q8) asks for the answer “in the form $a+b\ln c$ ”, and the repeated factor is the trap they set on purpose. The split is $\dfrac{1}{2x+1}-\dfrac{2}{x+2}+\dfrac{3}{(x+2)^2}$ , and only the last term breaks the $\ln$ pattern.

Collapsing to the required a + b ln c form

When the question fixes the shape of the answer, markers want the logs combined into one, not left as a sum.

Use the log laws on the bracket at the limits:

$A\ln m - A\ln n = A\ln\frac{m}{n}, \qquad b\ln m + b\ln n = b\ln(mn)$

Work out the bracket at both limits, subtract, then use $\ln m-\ln n=\ln\tfrac{m}{n}$ to fold everything into a single $a+b\ln c$ . This is just algebra plus integration, so write every line. The method marks are in the working, not the final number.

Common mistake

Three slips cost the marks: integrating the repeated factor as a log; not folding it down to the single

a+b\ln c

when asked; and number slips when you put the limits in. The quadratic-factor case also needs the

\ln

+ arctan split from §8.2, so do not forget the arctan half.

Your turn— tap to reveal the worked answer (9709/32 Nov 2022 Q10)

$f(x)=\dfrac{4-x+x^2}{(1+x)(2+x^2)}$ : (a) express in partial fractions, (b) find the exact value of $\int f(x)\,dx$ as a single logarithm. This is the quadratic-factor case. It splits as $\dfrac{2}{1+x}-\dfrac{x}{x^2+2}$ , and the second piece is a $\tfrac{f'}{f}$ form (the top is half the derivative of $x^2+2$ ):

Answer

\int f(x)\,dx = 2\ln(1+x) - \tfrac12\ln(x^2+2) + c

(9709/32 Nov 2022 Q10). Here the $x^2+2$ has no constant left on top after the split, so there is no arctan term, just two logs to combine into one.

The full case table: which piece becomes what

Once it is split, every rational integral in P3 is one of four shapes. Learn the right-hand column and integrating just becomes pattern-matching:

You see	It integrates to
$\dfrac{A}{x-p}$	$A\ln\|x-p\|$
$\dfrac{B}{(x-p)^2}$	$-\dfrac{B}{x-p}$ (power rule, not a log)
$\dfrac{Bx}{x^2+d}$	$\tfrac{B}{2}\ln(x^2+d)$
$\dfrac{C}{x^2+d}$	$\tfrac{C}{\sqrt d}\tan^{-1}\!\tfrac{x}{\sqrt d}$

A quadratic factor $\dfrac{Bx+C}{x^2+d}$ uses the last two rows together. You get an arctan only if a bare constant $C$ is left after the split.

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Integrating rational functions using partial fractions

This is the most common type in the chapter: partial fractions turn up in an integration question in about 30 of 42 recent papers.
The wording is almost always “express in partial fractions, hence find the exact value in the form $a+b\ln c$ .”
Get this right and you pick up the biggest single block of marks in the chapter.

Split first, then integrate one piece at a time

The algebra of splitting the fraction is Chapter 7 work, so take it as known. (Distinct linear → $\tfrac{A}{x-p}+\tfrac{B}{x-q}$ ; repeated linear → $\tfrac{A}{x-p}+\tfrac{B}{(x-p)^2}$ ; quadratic $cx^2+d$ → $\tfrac{Bx+C}{x^2+d}$ ; improper → divide first.)
This topic is about integrating each piece once it is split.

How to spot it: a factored bottom with a fraction on top → split into partial fractions (Ch 7), then integrate each piece one at a time.

Partial-fraction piece	Integrates to
$\dfrac{A}{x-p}$	$A\ln\|x-p\|$
$\dfrac{B}{(x-p)^2}$	$-\dfrac{B}{x-p}$ (power rule — NOT a log)
$\dfrac{Bx+C}{x^2+d}$	split into $\tfrac{B}{2}\ln(x^2+d)$ + $\tfrac{C}{\sqrt d}\tan^{-1}\!\frac{x}{\sqrt d}$

The main trap: a repeated factor is a reciprocal, not a log

Every linear factor on the bottom integrates to a $\ln$ , except the squared one.
A $\dfrac{B}{(x-p)^2}$ term is $B(x-p)^{-2}$ , and the power rule turns it into $-B(x-p)^{-1}=-\dfrac{B}{x-p}$ .
It is not a second log.

Common mistake

Integrating

\dfrac{B}{(x-p)^2}

B\ln|x-p|

is the most common slip the markscheme flags in this chapter. A repeated factor

(x-p)^2

on the bottom integrates to

-\dfrac{B}{x-p}

by the power rule.

Worked example

9709/31 Jun 2023: $f(x)=\dfrac{3-3x^2}{(2x+1)(x+2)^2}$

(a) Decompose (Ch 7 work) into $\dfrac{A}{2x+1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$ .
(b) Integrate term by term: the $\dfrac{A}{2x+1}$ → $\tfrac{A}{2}\ln|2x+1|$ , the $\dfrac{B}{x+2}$ → $B\ln|x+2|$ , and the repeated $\dfrac{C}{(x+2)^2}$ → $-\dfrac{C}{x+2}$ (reciprocal, not a log).
Put in the limits and collapse the logs into the required $a+b\ln c$ form using log laws.

Collapsing to the required a + b ln c form

When the question fixes the shape of the answer, markers want the logs combined into one, not left as a sum.

Use the log laws on the bracket at the limits:

$A\ln m - A\ln n = A\ln\frac{m}{n}, \qquad b\ln m + b\ln n = b\ln(mn)$

Common mistake

Three slips cost the marks: integrating the repeated factor as a log; not folding it down to the single

a+b\ln c

when asked; and number slips when you put the limits in. The quadratic-factor case also needs the

\ln

+ arctan split from §8.2, so do not forget the arctan half.

Your turn— tap to reveal the worked answer (9709/32 Nov 2022 Q10)

Answer

\int f(x)\,dx = 2\ln(1+x) - \tfrac12\ln(x^2+2) + c

(9709/32 Nov 2022 Q10). Here the $x^2+2$ has no constant left on top after the split, so there is no arctan term, just two logs to combine into one.

The full case table: which piece becomes what

Once it is split, every rational integral in P3 is one of four shapes. Learn the right-hand column and integrating just becomes pattern-matching:

You see	It integrates to
$\dfrac{A}{x-p}$	$A\ln\|x-p\|$
$\dfrac{B}{(x-p)^2}$	$-\dfrac{B}{x-p}$ (power rule, not a log)
$\dfrac{Bx}{x^2+d}$	$\tfrac{B}{2}\ln(x^2+d)$
$\dfrac{C}{x^2+d}$	$\tfrac{C}{\sqrt d}\tan^{-1}\!\tfrac{x}{\sqrt d}$

A quadratic factor $\dfrac{Bx+C}{x^2+d}$ uses the last two rows together. You get an arctan only if a bare constant $C$ is left after the split.