1.9.30 Integration using partial fractions (A-level only) - Revision Guide

What you'll learn

How to recognise when integration by partial fractions is appropriate.
How to decompose rational functions with distinct or repeated linear factors.
How to integrate the resulting fractions using logarithms.
How to handle improper fractions and definite integrals accurately.

Prerequisites

Rational functions

A rational function is a fraction in which both the numerator and denominator are polynomials, such as

3x+7(x−1)(x+2).\frac{3x+7}{(x-1)(x+2)}.

The degree of a polynomial is its highest power of $x$ . For example, $3x^2-5x+1$ has degree 2.

A rational function is proper if the degree of its numerator is smaller than the degree of its denominator. Otherwise, it is improper.

Definition

Partial fractions

A partial-fraction decomposition rewrites one rational function as a sum of simpler rational functions. These simpler fractions can often be integrated directly.

For example, an expression with two distinct linear factors may have the form

3x+7(x−1)(x+2)=Ax−1+Bx+2,\frac{3x+7}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2},

where $A$ and $B$ are constants to be found.

The logarithmic integral

The key standard result is

dx=ln⁡∣x∣+C.\int \frac{1}{x}\,dx=\ln|x|+C.

More generally,

dx=1aln⁡∣ax+b∣+C.\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b|+C.

The factor $1a\frac{1}{a}$ appears because the derivative of $a x + b$ is $a$ .

Example

Integrating a linear denominator

Evaluate $dx\displaystyle \int \frac{5}{3x-4}\,dx$ .

Use the standard form with $a = 3$ , so integrating $13x−4\frac{1}{3x-4}$ introduces a factor of $13\frac{1}{3}$ .
Multiply this by the numerator 5:
$dx=53ln⁡∣3x−4∣+C.\int \frac{5}{3x-4}\,dx = \frac{5}{3}\ln|3x-4|+C.$
Differentiate to check:
$ddx(53ln⁡∣3x−4∣)=53⋅33x−4=53x−4.\frac{d}{dx}\left(\frac{5}{3}\ln|3x-4|\right) = \frac{5}{3}\cdot\frac{3}{3x-4} = \frac{5}{3x-4}.$

Common Mistake

Forgetting the inner derivative

In $dx\int \frac{1}{ax+b}\,dx$ , do not write simply $ln⁡∣ax+b∣\ln|ax+b|$ . You must divide by the coefficient $a$ .

Distinct linear factors

A linear factor has the form $a x + b$ . If the denominator contains different linear factors, assign one constant numerator to each factor.

For example,

px+q(x−a)(x−b)=Ax−a+Bx−b,a≠b.\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}, \qquad a\neq b.

You can find $A$ and $B$ by multiplying through by the original denominator:

p x + q = A (x - b) + B (x - a) .

You may then substitute convenient values of $x$ , usually the values that make individual factors zero.

Key Idea

Decompose before integrating

Partial fractions change one difficult rational integral into a sum of standard integrals. Find the decomposition first, then integrate each term separately.

Example

Integrating with two distinct linear factors

Evaluate

dx.\int \frac{5x+1}{(x-1)(x+2)}\,dx.

Write the correct decomposition:
$5x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}.$
Multiply by $(x - 1) (x + 2)$ :
$5 x + 1 = A (x + 2) + B (x - 1) .$
Substitute $x = 1$ to eliminate the term containing $B$ :
$\quad\Rightarrow\quad A=2.$
Substitute $x = - 2$ to eliminate the term containing $A$ :
$\quad\Rightarrow\quad B=3.$
Integrate the decomposition:
$dx=2ln⁡∣x−1∣+3ln⁡∣x+2∣+C.\begin{aligned} \int \frac{5x+1}{(x-1)(x+2)}\,dx &= \int\left(\frac{2}{x-1}+\frac{3}{x+2}\right)\,dx\\ &= 2\ln|x-1|+3\ln|x+2|+C. \end{aligned}$

Tip

Use the roots strategically

After clearing the denominator, substitute the root of each linear factor. This makes one term survive while the others become zero.

Linear factors with coefficients

A denominator factor might be $2 x + 1$ rather than $x + 1$ . The decomposition method is unchanged, but the coefficient of $x$ matters when you integrate.

Example

Accounting for linear coefficients

Evaluate

dx.\int \frac{7x+1}{(2x+1)(x-1)}\,dx.

Decompose the fraction:
$7x+1(2x+1)(x−1)=A2x+1+Bx−1.\frac{7x+1}{(2x+1)(x-1)} = \frac{A}{2x+1}+\frac{B}{x-1}.$
Clear the denominator:
$7 x + 1 = A (x - 1) + B (2 x + 1) .$
Substitute $x = 1$ :
$\quad\Rightarrow\quad B=\frac{8}{3}.$
Substitute $x=−12x=-\frac{1}{2}$ :
$−52=−32A⇒A=53.-\frac{5}{2}=-\frac{3}{2}A \quad\Rightarrow\quad A=\frac{5}{3}.$
Integrate, remembering that the derivative of $2 x + 1$ is 2:
$dx=56ln⁡∣2x+1∣+83ln⁡∣x−1∣+C.\begin{aligned} \int\left(\frac{5}{3(2x+1)}+\frac{8}{3(x-1)}\right)\,dx &= \frac{5}{6}\ln|2x+1| +\frac{8}{3}\ln|x-1|+C. \end{aligned}$

Repeated linear factors

A factor is repeated when it occurs more than once, as in $x-2)^2$ .

You must include a fraction for every power of the repeated factor:

P(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b.\frac{P(x)}{(x-a)^2(x-b)} = \frac{A}{x-a} +\frac{B}{(x-a)^2} +\frac{C}{x-b}.

Leaving out one of these terms gives an incomplete decomposition.

Example

Integrating with a repeated linear factor

Evaluate

dx.\int \frac{3x+5}{x(x+1)^2}\,dx.

Include every required partial fraction:
$3x+5x(x+1)2=Ax+Bx+1+C(x+1)2.\frac{3x+5}{x(x+1)^2} = \frac{A}{x} +\frac{B}{x+1} +\frac{C}{(x+1)^2}.$
Multiply through by $x(x+1)^2$ :
$3x+5=A(x+1)^2+Bx(x+1)+Cx.$
Substitute $x = 0$ to obtain $A = 5$ , and substitute $x = - 1$ to obtain $C = - 2$ .
Compare the coefficients of $x^2$ . The left side has coefficient zero, so
$\quad\Rightarrow\quad B=-5.$
Integrate each term, writing $x+1)^{-2}$ in power form:
$dx=5ln⁡∣x∣−5ln⁡∣x+1∣+2x+1+C.\begin{aligned} \int\left(\frac{5}{x}-\frac{5}{x+1}-\frac{2}{(x+1)^2}\right)\,dx &= 5\ln|x|-5\ln|x+1|+\frac{2}{x+1}+C. \end{aligned}$

Common Mistake

Treating every term as a logarithm

Only a denominator to the first power produces a logarithm directly. Integrate $1(x−a)2\frac{1}{(x-a)^2}$ using the power rule.

Improper rational functions

Partial-fraction decomposition requires a proper rational function. If the numerator has degree greater than or equal to the denominator, first use polynomial division.

Example

Dividing before decomposition

Evaluate

dx.\int \frac{x^2+2x+3}{x(x+1)}\,dx.

Since the numerator and denominator both have degree 2, divide first:
$x2+2x+3x2+x=1+x+3x(x+1).\frac{x^2+2x+3}{x^2+x} = 1+\frac{x+3}{x(x+1)}.$
Decompose the proper remainder:
$x+3x(x+1)=Ax+Bx+1.\frac{x+3}{x(x+1)} = \frac{A}{x}+\frac{B}{x+1}.$
Clearing the denominator gives
$x + 3 = A (x + 1) + B x .$
Substituting $x = 0$ gives $A = 3$ . Comparing coefficients of $x$ then gives $1 = A + B$ , so $B = - 2$ .
Integrate the complete expression:
$dx=x+3ln⁡∣x∣−2ln⁡∣x+1∣+C.\begin{aligned} \int \frac{x^2+2x+3}{x(x+1)}\,dx &= \int\left(1+\frac{3}{x}-\frac{2}{x+1}\right)\,dx\\ &= x+3\ln|x|-2\ln|x+1|+C. \end{aligned}$

Definite integrals

For a definite integral, find an antiderivative and evaluate it at the limits. Check that the original integrand is defined throughout the interval.

Common Mistake

Discontinuities inside the interval

If a denominator is zero between the limits, the ordinary definite integral is improper and may not exist. Do not substitute into a logarithmic antiderivative without checking the interval.

Example

Evaluating a definite integral

Evaluate

dx.\int_2^3 \frac{1}{x(x+1)}\,dx.

Decompose the integrand:
$1x(x+1)=1x−1x+1.\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}.$
Integrate and apply the limits:
$dx=[ln⁡∣x∣−ln⁡∣x+1∣]23.\int_2^3 \frac{1}{x(x+1)}\,dx = \left[\ln|x|-\ln|x+1|\right]_2^3.$
Substitute the upper and lower limits:
$(ln⁡3−ln⁡4)−(ln⁡2−ln⁡3)=ln⁡(98).\begin{aligned} \left(\ln 3-\ln 4\right) -\left(\ln 2-\ln 3\right) &= \ln\left(\frac{9}{8}\right). \end{aligned}$

Checking your result

You can check a decomposition by recombining the fractions over a common denominator. You can check the final integral by differentiating it.

Keep absolute-value signs in indefinite logarithmic integrals unless the domain guarantees that the expression inside the logarithm is positive.

Exam technique

In the exam

Check that the rational function is proper; use polynomial division first if it is not.
Factorise the denominator fully and include one partial fraction for every distinct factor and every power of a repeated factor.
Clear the denominator, then use convenient substitutions and coefficient comparison to find the constants.
Integrate each term carefully, allowing for the derivative of each linear denominator.
Use absolute values in logarithms and check for denominator zeros before evaluating a definite integral.

Self review

Check yourself

How would you decompose $4x+1(x−2)(x+3)\frac{4x+1}{(x-2)(x+3)}$ before integrating?
Which partial-fraction terms are required when the denominator is $x+1)^3(x-4)$ ?
Why must polynomial division come before partial fractions for $x3+1x2−x\frac{x^3+1}{x^2-x}$ ?

What you'll learn

How to recognise when integration by partial fractions is appropriate.
How to decompose rational functions with distinct or repeated linear factors.
How to integrate the resulting fractions using logarithms.
How to handle improper fractions and definite integrals accurately.

Prerequisites

Rational functions

A rational function is a fraction in which both the numerator and denominator are polynomials, such as

3x+7(x−1)(x+2).\frac{3x+7}{(x-1)(x+2)}.

The degree of a polynomial is its highest power of $x$ . For example, $3x^2-5x+1$ has degree 2.

A rational function is proper if the degree of its numerator is smaller than the degree of its denominator. Otherwise, it is improper.

Definition

Partial fractions

A partial-fraction decomposition rewrites one rational function as a sum of simpler rational functions. These simpler fractions can often be integrated directly.

For example, an expression with two distinct linear factors may have the form

3x+7(x−1)(x+2)=Ax−1+Bx+2,\frac{3x+7}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2},

where $A$ and $B$ are constants to be found.

The logarithmic integral

The key standard result is

dx=ln⁡∣x∣+C.\int \frac{1}{x}\,dx=\ln|x|+C.

More generally,

dx=1aln⁡∣ax+b∣+C.\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b|+C.

The factor $1a\frac{1}{a}$ appears because the derivative of $a x + b$ is $a$ .

Example

Integrating a linear denominator

Evaluate $dx\displaystyle \int \frac{5}{3x-4}\,dx$ .

Use the standard form with $a = 3$ , so integrating $13x−4\frac{1}{3x-4}$ introduces a factor of $13\frac{1}{3}$ .
Multiply this by the numerator 5:
$dx=53ln⁡∣3x−4∣+C.\int \frac{5}{3x-4}\,dx = \frac{5}{3}\ln|3x-4|+C.$
Differentiate to check:
$ddx(53ln⁡∣3x−4∣)=53⋅33x−4=53x−4.\frac{d}{dx}\left(\frac{5}{3}\ln|3x-4|\right) = \frac{5}{3}\cdot\frac{3}{3x-4} = \frac{5}{3x-4}.$

Common Mistake

Forgetting the inner derivative

In $dx\int \frac{1}{ax+b}\,dx$ , do not write simply $ln⁡∣ax+b∣\ln|ax+b|$ . You must divide by the coefficient $a$ .

Distinct linear factors

A linear factor has the form $a x + b$ . If the denominator contains different linear factors, assign one constant numerator to each factor.

For example,

px+q(x−a)(x−b)=Ax−a+Bx−b,a≠b.\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}, \qquad a\neq b.

You can find $A$ and $B$ by multiplying through by the original denominator:

p x + q = A (x - b) + B (x - a) .

You may then substitute convenient values of $x$ , usually the values that make individual factors zero.

Key Idea

Decompose before integrating

Partial fractions change one difficult rational integral into a sum of standard integrals. Find the decomposition first, then integrate each term separately.

Example

Integrating with two distinct linear factors

Evaluate

dx.\int \frac{5x+1}{(x-1)(x+2)}\,dx.

Write the correct decomposition:
$5x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}.$
Multiply by $(x - 1) (x + 2)$ :
$5 x + 1 = A (x + 2) + B (x - 1) .$
Substitute $x = 1$ to eliminate the term containing $B$ :
$\quad\Rightarrow\quad A=2.$
Substitute $x = - 2$ to eliminate the term containing $A$ :
$\quad\Rightarrow\quad B=3.$
Integrate the decomposition:
$dx=2ln⁡∣x−1∣+3ln⁡∣x+2∣+C.\begin{aligned} \int \frac{5x+1}{(x-1)(x+2)}\,dx &= \int\left(\frac{2}{x-1}+\frac{3}{x+2}\right)\,dx\\ &= 2\ln|x-1|+3\ln|x+2|+C. \end{aligned}$

Tip

Use the roots strategically

After clearing the denominator, substitute the root of each linear factor. This makes one term survive while the others become zero.

Linear factors with coefficients

A denominator factor might be $2 x + 1$ rather than $x + 1$ . The decomposition method is unchanged, but the coefficient of $x$ matters when you integrate.

Example

Accounting for linear coefficients

Evaluate

dx.\int \frac{7x+1}{(2x+1)(x-1)}\,dx.

Decompose the fraction:
$7x+1(2x+1)(x−1)=A2x+1+Bx−1.\frac{7x+1}{(2x+1)(x-1)} = \frac{A}{2x+1}+\frac{B}{x-1}.$
Clear the denominator:
$7 x + 1 = A (x - 1) + B (2 x + 1) .$
Substitute $x = 1$ :
$\quad\Rightarrow\quad B=\frac{8}{3}.$
Substitute $x=−12x=-\frac{1}{2}$ :
$−52=−32A⇒A=53.-\frac{5}{2}=-\frac{3}{2}A \quad\Rightarrow\quad A=\frac{5}{3}.$
Integrate, remembering that the derivative of $2 x + 1$ is 2:
$dx=56ln⁡∣2x+1∣+83ln⁡∣x−1∣+C.\begin{aligned} \int\left(\frac{5}{3(2x+1)}+\frac{8}{3(x-1)}\right)\,dx &= \frac{5}{6}\ln|2x+1| +\frac{8}{3}\ln|x-1|+C. \end{aligned}$

Repeated linear factors

A factor is repeated when it occurs more than once, as in $x-2)^2$ .

You must include a fraction for every power of the repeated factor:

P(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b.\frac{P(x)}{(x-a)^2(x-b)} = \frac{A}{x-a} +\frac{B}{(x-a)^2} +\frac{C}{x-b}.

Leaving out one of these terms gives an incomplete decomposition.

Example

Integrating with a repeated linear factor

Evaluate

dx.\int \frac{3x+5}{x(x+1)^2}\,dx.

Include every required partial fraction:
$3x+5x(x+1)2=Ax+Bx+1+C(x+1)2.\frac{3x+5}{x(x+1)^2} = \frac{A}{x} +\frac{B}{x+1} +\frac{C}{(x+1)^2}.$
Multiply through by $x(x+1)^2$ :
$3x+5=A(x+1)^2+Bx(x+1)+Cx.$
Substitute $x = 0$ to obtain $A = 5$ , and substitute $x = - 1$ to obtain $C = - 2$ .
Compare the coefficients of $x^2$ . The left side has coefficient zero, so
$\quad\Rightarrow\quad B=-5.$
Integrate each term, writing $x+1)^{-2}$ in power form:
$dx=5ln⁡∣x∣−5ln⁡∣x+1∣+2x+1+C.\begin{aligned} \int\left(\frac{5}{x}-\frac{5}{x+1}-\frac{2}{(x+1)^2}\right)\,dx &= 5\ln|x|-5\ln|x+1|+\frac{2}{x+1}+C. \end{aligned}$

Common Mistake

Treating every term as a logarithm

Only a denominator to the first power produces a logarithm directly. Integrate $1(x−a)2\frac{1}{(x-a)^2}$ using the power rule.

Improper rational functions

Partial-fraction decomposition requires a proper rational function. If the numerator has degree greater than or equal to the denominator, first use polynomial division.

Example

Dividing before decomposition

Evaluate

dx.\int \frac{x^2+2x+3}{x(x+1)}\,dx.

Since the numerator and denominator both have degree 2, divide first:
$x2+2x+3x2+x=1+x+3x(x+1).\frac{x^2+2x+3}{x^2+x} = 1+\frac{x+3}{x(x+1)}.$
Decompose the proper remainder:
$x+3x(x+1)=Ax+Bx+1.\frac{x+3}{x(x+1)} = \frac{A}{x}+\frac{B}{x+1}.$
Clearing the denominator gives
$x + 3 = A (x + 1) + B x .$
Substituting $x = 0$ gives $A = 3$ . Comparing coefficients of $x$ then gives $1 = A + B$ , so $B = - 2$ .
Integrate the complete expression:
$dx=x+3ln⁡∣x∣−2ln⁡∣x+1∣+C.\begin{aligned} \int \frac{x^2+2x+3}{x(x+1)}\,dx &= \int\left(1+\frac{3}{x}-\frac{2}{x+1}\right)\,dx\\ &= x+3\ln|x|-2\ln|x+1|+C. \end{aligned}$

Definite integrals

For a definite integral, find an antiderivative and evaluate it at the limits. Check that the original integrand is defined throughout the interval.

Common Mistake

Discontinuities inside the interval

If a denominator is zero between the limits, the ordinary definite integral is improper and may not exist. Do not substitute into a logarithmic antiderivative without checking the interval.

Example

Evaluating a definite integral

Evaluate

dx.\int_2^3 \frac{1}{x(x+1)}\,dx.

Decompose the integrand:
$1x(x+1)=1x−1x+1.\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}.$
Integrate and apply the limits:
$dx=[ln⁡∣x∣−ln⁡∣x+1∣]23.\int_2^3 \frac{1}{x(x+1)}\,dx = \left[\ln|x|-\ln|x+1|\right]_2^3.$
Substitute the upper and lower limits:
$(ln⁡3−ln⁡4)−(ln⁡2−ln⁡3)=ln⁡(98).\begin{aligned} \left(\ln 3-\ln 4\right) -\left(\ln 2-\ln 3\right) &= \ln\left(\frac{9}{8}\right). \end{aligned}$

Checking your result

You can check a decomposition by recombining the fractions over a common denominator. You can check the final integral by differentiating it.

Keep absolute-value signs in indefinite logarithmic integrals unless the domain guarantees that the expression inside the logarithm is positive.

Exam technique

In the exam

Check that the rational function is proper; use polynomial division first if it is not.
Factorise the denominator fully and include one partial fraction for every distinct factor and every power of a repeated factor.
Clear the denominator, then use convenient substitutions and coefficient comparison to find the constants.
Integrate each term carefully, allowing for the derivative of each linear denominator.
Use absolute values in logarithms and check for denominator zeros before evaluating a definite integral.

Self review

Check yourself

How would you decompose $4x+1(x−2)(x+3)\frac{4x+1}{(x-2)(x+3)}$ before integrating?
Which partial-fraction terms are required when the denominator is $x+1)^3(x-4)$ ?
Why must polynomial division come before partial fractions for $x3+1x2−x\frac{x^3+1}{x^2-x}$ ?

1.11.6 Integration using partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions

The logarithmic integral

Distinct linear factors

Linear factors with coefficients

Repeated linear factors

Improper rational functions

Definite integrals

Checking your result

Review 1.9.30 Integration using partial fractions (A-level only) by teaching Genie

1.11.6 Integration using partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions

The logarithmic integral

Distinct linear factors

Linear factors with coefficients

Repeated linear factors

Improper rational functions

Definite integrals

Checking your result

Review 1.9.30 Integration using partial fractions (A-level only) by teaching Genie

1.11.6 Integration using partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions

The logarithmic integral

Distinct linear factors

Linear factors with coefficients

Repeated linear factors

Improper rational functions

Definite integrals

Checking your result

Review 1.9.30 Integration using partial fractions (A-level only) by teaching Genie

1.9.30 Integration using partial fractions (A-level only) Revision Guide

1.11.6 Integration using partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions

The logarithmic integral

Distinct linear factors

Linear factors with coefficients

Repeated linear factors

Improper rational functions

Definite integrals

Checking your result

Review 1.9.30 Integration using partial fractions (A-level only) by teaching Genie

1.9.30 Integration using partial fractions (A-level only) Revision Guide