What is 8.6: Integrating with partial fractions?

Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants. For A/x-a, the integral is A|x-a|+C; retain absolute values unless the domain makes the sign known.

How many marks are 8.6 questions worth?

8.6 practice drills on this page are worth 3–7 marks each; marks in a real exam vary by question.

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8.6

Integrate using partial fractions that are linear in the denominator.

Draft — not yet indexed

Integrating with partial fractions

Worked answers and methods for 8.6 on Edexcel A-level Maths 9MA0.

Explanation

Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants. For $\frac{A}{x-a}$ , the integral is $A\ln|x-a|+C$ ; retain absolute values unless the domain makes the sign known.
Determine constants by equating coefficients or substituting convenient values that make all but one term vanish.
A common error is to integrate the unfactorised denominator as though $\int f'(x)/f(x)\,\mathrm dx$ applied when the numerator is not its derivative.
Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$.
Then $5x+1=A(x+2)+B(x-1)$ .

Worked example

Express $\frac{5x+1}{(x-1)(x+2)}$ in partial fractions and hence integrate it.

1.Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$ .
2.Then $5x+1=A(x+2)+B(x-1)$ .
3.Substituting $x=1$ gives $A=2$ , and $x=-2$ gives $B=3$ .
4.Integrating the decomposition gives $2\ln|x-1|+3\ln|x+2|+C$ .

Answer: $\frac{5x+1}{(x-1)(x+2)}=\frac2{x-1}+\frac3{x+2}$ ; $2\ln|x-1|+3\ln|x+2|+C$

Common mistakes

Don't drop the absolute-value signs after integrating a linear denominator to a logarithm.
Don't split the fraction without solving the coefficient identity, so the logarithmic coefficients are incorrect.

Exam tip

Find partial-fraction constants before integrating and preserve absolute-value signs in logarithmic antiderivatives.

Worked practice

Tier 1 · Easy

Find

\int\left(\frac3{x-1}+\frac2{x+2}\right)\,\mathrm dx

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$3\ln\|x-1\|+2\ln\|x+2\|+C$	3
Notes
Integrate each linear-denominator term directly: $\int3/(x-1)\,\mathrm dx=3\ln\|x-1\|$ and $\int2/(x+2)\,\mathrm dx=2\ln\|x+2\|$ . Add $C$ .

(3 marks)

Tier 2 · Standard

Express

\dfrac{2x^2+3x-1}{x(x+1)}

in the form

A+\dfrac{B}{x}+\dfrac{C}{x+1}

and hence integrate it.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac{2x^2+3x-1}{x(x+1)}=2-\dfrac1x+\dfrac{2}{x+1}$ $2x-\ln\|x\|+2\ln\|x+1\|+C$	5
Notes
First divide: $(2x^2+3x-1)/(x^2+x)=2+(x-1)/[x(x+1)]$ . Write $(x-1)/[x(x+1)]=A/x+B/(x+1)$ , so $x-1=(A+B)x+A$ . Hence $A=-1$ and $B=2$ . Integrating $2-1/x+2/(x+1)$ gives $2x-\ln\|x\|+2\ln\|x+1\|+C$ .

(5 marks)

Tier 3 · Hard

Evaluate exactly

\int_0^1\frac{5x+1}{(x+1)(x+2)}\,\mathrm dx

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$9\ln3-13\ln2$	7
Notes
Write $\frac{5x+1}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}$ . Then $5x+1=A(x+2)+B(x+1)$ , giving $A=-4$ and $B=9$ . The integral is $[-4\ln(x+1)+9\ln(x+2)]_0^1=(-4\ln2+9\ln3)-9\ln2=9\ln3-13\ln2$ .

(7 marks)

Tier 1 · Easy

Given that

\dfrac{1}{(x+1)(x+3)}=\dfrac{1}{2(x+1)}-\dfrac{1}{2(x+3)}

, integrate

\dfrac{1}{(x+1)(x+3)}

and combine the logarithms.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac12\ln\left\|\dfrac{x+1}{x+3}\right\|+C$	3
Notes
Integrating the supplied terms gives $\frac12\ln\|x+1\|-\frac12\ln\|x+3\|+C$ . Using the logarithm laws, this is $\frac12\ln\|\frac{x+1}{x+3}\|+C$ .

(3 marks)

Tier 2 · Standard

Express

\dfrac{4x-7}{(x-2)(x+3)}

in partial fractions and hence integrate it.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\dfrac{4x-7}{(x-2)(x+3)}=\dfrac{1}{5(x-2)}+\dfrac{19}{5(x+3)}$ $\dfrac15\ln\|x-2\|+\dfrac{19}{5}\ln\|x+3\|+C$	4
Notes
Write the fraction as $A/(x-2)+B/(x+3)$ . Then $4x-7=A(x+3)+B(x-2)$ . Substituting $x=2$ gives $A=1/5$ , and substituting $x=-3$ gives $B=19/5$ . Integrating the decomposition gives the stated logarithmic result.

(4 marks)

Tier 3 · Hard

For

x>0

, a function

F

satisfies

F'(x)=\dfrac{3}{(x+2)(x+5)}

and

F(1)=0

. (a) Find

F(x)

. (b) Hence solve

F(x)=\ln\left(\dfrac{10}{7}\right)

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$F(x)=\ln\left(\dfrac{2(x+2)}{x+5}\right)$ $x=\dfrac{11}{2}$	6
Notes
$3/[(x+2)(x+5)]=1/(x+2)-1/(x+5)$ , so $F(x)=\ln(x+2)-\ln(x+5)+C$ . From $F(1)=0$ , $C=\ln2$ , giving $F(x)=\ln[2(x+2)/(x+5)]$ . Equating logarithm arguments gives $2(x+2)/(x+5)=10/7$ , so $14x+28=10x+50$ and $x=11/2$ .

(6 marks)

Tier 2 · Standard

Express

\dfrac{4x+7}{(x+1)(2x+3)}

in partial fractions and hence integrate it.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac{4x+7}{(x+1)(2x+3)}=\dfrac3{x+1}-\dfrac2{2x+3}$ $3\ln\|x+1\|-\ln\|2x+3\|+C$	5
Notes
Write the fraction as $A/(x+1)+B/(2x+3)$ . Then $4x+7=A(2x+3)+B(x+1)$ . Equating coefficients gives $2A+B=4$ and $3A+B=7$ , so $A=3$ and $B=-2$ . Integrating, including the inner coefficient in the second logarithm, gives $3\ln\|x+1\|-\ln\|2x+3\|+C$ .

(5 marks)

Tier 3 · Hard

Evaluate exactly

\displaystyle\int_1^2\dfrac{4x^2+7x+3}{(2x-1)(x+2)}\,\mathrm dx

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$2+\dfrac52\ln3-\ln4$	6
Notes
Division and partial fractions give $\frac{4x^2+7x+3}{(2x-1)(x+2)}=2+\frac3{2x-1}-\frac1{x+2}$ . An antiderivative is $2x+\frac32\ln\|2x-1\|-\ln\|x+2\|$ . Evaluation from $1$ to $2$ gives $4+\frac32\ln3-\ln4-(2-\ln3)=2+\frac52\ln3-\ln4$ .

(6 marks)

Tier 3 · Hard

Given that

\dfrac{9x-8}{(x-4)(3x+2)}=\dfrac{A}{x-4}+\dfrac{B}{3x+2}

, find

A

and

B

. Hence, for

x>5

, find

\displaystyle\int_5^x\dfrac{9t-8}{(t-4)(3t+2)}\,\mathrm dt

in the form

\ln f(x)

. Solve

\displaystyle\int_5^x\dfrac{9t-8}{(t-4)(3t+2)}\,\mathrm dt=\ln\left(\dfrac{80}{17}\right)

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$A=2$ , $B=3$ $\displaystyle\ln\left(\dfrac{(x-4)^2(3x+2)}{17}\right)$ $x=6$	6
Notes
Multiplying by $(x-4)(3x+2)$ gives $9x-8=A(3x+2)+B(x-4)$ . Substituting $x=4$ gives $A=2$ , and substituting $x=-2/3$ gives $B=3$ . An antiderivative is $2\ln(t-4)+\ln(3t+2)$ for $t>5$ . Applying the limits gives $\ln[(x-4)^2(3x+2)/17]$ . Equating logarithm arguments gives $(x-4)^2(3x+2)=80$ . The left-hand side is strictly increasing for $x>5$ , and $x=6$ satisfies the equation, so $x=6$ is the unique solution.

(6 marks)

Q10

Tier 3 · Hard

10.

Express

\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}

in partial fractions and hence evaluate exactly

\displaystyle\int_0^1\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}\,\mathrm dx

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}=\dfrac1{x+1}+\dfrac2{x+2}-\dfrac1{x+3}$ $3\ln\left(\dfrac32\right)$	6
Notes
Writing the fraction as $A/(x+1)+B/(x+2)+C/(x+3)$ and substituting $x=-1,-2,-3$ gives $A=1$ , $B=2$ and $C=-1$ . An antiderivative is $\ln\|x+1\|+2\ln\|x+2\|-\ln\|x+3\|$ . Evaluation from $0$ to $1$ gives $3\ln3-3\ln2=3\ln(3/2)$ .

(6 marks)

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