Integrating with Partial Fractions

A2 · P3 · 14 min

A rational function such as $\dfrac{4}{x(x + 2)}$ cannot be integrated as it stands: it is not a standard form, and splitting the denominator is not allowed. Once it is written in partial fractions, every piece is standard: a logarithm, a reciprocal, or a logarithm plus an inverse tangent. The syllabus asks you to "integrate rational functions by means of decomposition into partial fractions", restricted to the denominators you met in the algebra topic. On P3 this is a classic two-part question worth about eight marks, and it also appears inside differential equations, where separating the variables produces a fraction that needs splitting.

Why partial fractions make integration possible

Adding $\dfrac{2}{x}$ and $\dfrac{-2}{x + 2}$ gives $\dfrac{4}{x(x + 2)}$ . Integrating the combined fraction directly looks impossible, but each of the two simple fractions integrates at once to a logarithm:

\int\frac{4}{x(x + 2)}\, dx = \int\left(\frac{2}{x} - \frac{2}{x + 2}\right) dx = 2\ln\lvert x\rvert - 2\ln\lvert x + 2\rvert + c = 2\ln\left\lvert\frac{x}{x + 2}\right\rvert + c

That is the whole method: decompose, integrate each piece, tidy the logarithms.

The pieces and their integrals

The syllabus restricts denominators to the types $(ax + b)(cx + d)(ex + f)$ , $(ax + b)(cx + d)^2$ and $(ax + b)(cx^2 + d)$ , with numerator degree at most equal to the denominator's. After decomposition, only four kinds of term can appear.

Integrating each kind of partial fraction

Term	Integral	Why
$K$ (from an improper fraction)	$Kx$	constant
$\dfrac{A}{ax + b}$	$\dfrac{A}{a}\ln\lvert ax + b\rvert$	linear inside, power $-1$
$\dfrac{C}{(cx + d)^2}$	$-\dfrac{C}{c(cx + d)}$	power $-2$ , not a logarithm
$\dfrac{Bx + C}{x^2 + k^2}$	$\dfrac{B}{2}\ln(x^2 + k^2) + \dfrac{C}{k}\tan^{-1}\left(\dfrac{x}{k}\right)$	split the numerator

(each $+\,c$ for an indefinite integral)

Each line is a result from an earlier note:

Linear factors give logarithms. Remember to divide by the coefficient of $x$ : $\displaystyle\int\frac{3}{2x - 1}\, dx = \tfrac{3}{2}\ln\lvert 2x - 1\rvert + c$ .
A repeated factor gives one logarithm (from $\dfrac{B}{cx + d}$ ) and one reciprocal (from $\dfrac{C}{(cx + d)^2}$ ). Write the squared term as $C(cx + d)^{-2}$ and use the power rule: $\displaystyle\int C(cx + d)^{-2}\, dx = \frac{C(cx + d)^{-1}}{-1 \times c} + c$ . Students who write a logarithm here lose the marks.
A quadratic factor $x^2 + k^2$ gives a logarithm and an inverse tangent after splitting the numerator, exactly as in Integrating f'(x)/f(x):

\frac{Bx + C}{x^2 + k^2} = \frac{B}{2}\cdot\frac{2x}{x^2 + k^2} + C\cdot\frac{1}{x^2 + k^2}

If the quadratic is $cx^2 + d$ with $c \ne 1$ , the logarithm part has $\dfrac{B}{2c}$ and the inverse tangent needs the coefficient taken out first.

Integrating a rational function

Check the degrees. If numerator and denominator have the same degree, the decomposition includes a constant $K$ .
Factorise the denominator and write the partial fraction form.
Find the constants (substitute the roots of the linear factors, then compare coefficients). Check with a spare value of $x$ .
Integrate each term using the table, dividing by the coefficient of $x$ in each linear factor.
For a definite integral, substitute both limits and combine the logarithms into a single logarithm, or the form the question asks for.

Tidying logarithms

Answers are usually requested in a form such as $\ln a$ , $a + \ln b$ or $\ln\dfrac{p}{q}$ . Use

p\ln a + q\ln b = \ln\left(a^p b^q\right), \qquad p\ln a - q\ln b = \ln\frac{a^p}{b^q}

Collect all the logarithms at one limit before subtracting the other, or collect terms with the same argument first; either way, write the intermediate line so the examiner can follow it. For instance $2\ln 2 - 2\ln 4 + 2\ln 3 = 2\ln\dfrac{2 \times 3}{4} = 2\ln\dfrac{3}{2} = \ln\dfrac{9}{4}$ .

Worked examples

Routine: two linear factors

Find $\displaystyle\int\frac{7x + 4}{(x + 2)(2x - 1)}\, dx$ .

Solution

Write $\dfrac{7x + 4}{(x + 2)(2x - 1)} \equiv \dfrac{A}{x + 2} + \dfrac{B}{2x - 1}$ , so $7x + 4 \equiv A(2x - 1) + B(x + 2)$ .

$x = -2$ : $-10 = -5A$ , so $A = 2$ . $\quad x = \tfrac{1}{2}$ : $\tfrac{15}{2} = \tfrac{5}{2}B$ , so $B = 3$ .

\int\left(\frac{2}{x + 2} + \frac{3}{2x - 1}\right) dx = 2\ln\lvert x + 2\rvert + \tfrac{3}{2}\ln\lvert 2x - 1\rvert + c

A definite integral as a single logarithm

Show that $\displaystyle\int_1^2\frac{4}{x(x + 2)}\, dx = \ln\frac{9}{4}$ .

Solution

$\dfrac{4}{x(x + 2)} \equiv \dfrac{A}{x} + \dfrac{B}{x + 2}$ , so $4 \equiv A(x + 2) + Bx$ . $x = 0$ : $A = 2$ . $x = -2$ : $B = -2$ .

\int_1^2\left(\frac{2}{x} - \frac{2}{x + 2}\right) dx = \Big[2\ln x - 2\ln(x + 2)\Big]_1^2 = (2\ln 2 - 2\ln 4) - (2\ln 1 - 2\ln 3)

= 2\ln 2 - 2\ln 4 + 2\ln 3 = 2\ln\frac{2 \times 3}{4} = 2\ln\frac{3}{2} = \ln\frac{9}{4}

y = 4 / (x (x + 2)) fill 1 2 y = 4 / (x (x + 2))

The shaded area under $y = \dfrac{4}{x(x + 2)}$ between $x = 1$ and $x = 2$ is $\ln\tfrac{9}{4} \approx 0.811$ .

A repeated factor

(a) Express $\dfrac{4}{(x + 3)(x + 1)^2}$ in partial fractions.

(b) Hence show that $\displaystyle\int_0^1\frac{4}{(x + 3)(x + 1)^2}\, dx = 1 + \ln\frac{2}{3}$ .

Solution

(a) $\dfrac{4}{(x + 3)(x + 1)^2} \equiv \dfrac{A}{x + 3} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$ , so

4 \equiv A(x + 1)^2 + B(x + 3)(x + 1) + C(x + 3)

$x = -3$ : $4 = 4A$ , $A = 1$ . $\quad x = -1$ : $4 = 2C$ , $C = 2$ . $\quad$ Coefficient of $x^2$ : $0 = A + B$ , so $B = -1$ .

Check with $x = 0$ : $A + 3B + 3C = 1 - 3 + 6 = 4$ . Correct.

\frac{4}{(x + 3)(x + 1)^2} = \frac{1}{x + 3} - \frac{1}{x + 1} + \frac{2}{(x + 1)^2}

(b) The last term is $2(x + 1)^{-2}$ , which integrates to $-2(x + 1)^{-1}$ :

\int_0^1 \ldots\, dx = \Big[\ln(x + 3) - \ln(x + 1) - \frac{2}{x + 1}\Big]_0^1 = (\ln 4 - \ln 2 - 1) - (\ln 3 - \ln 1 - 2)

= \ln 2 - \ln 3 + 1 = 1 + \ln\frac{2}{3}

A quadratic factor

(a) Express $\dfrac{3x^2 - 3x + 4}{(x + 1)(x^2 + 4)}$ in partial fractions.

(b) Hence find the exact value of $\displaystyle\int_0^2\frac{3x^2 - 3x + 4}{(x + 1)(x^2 + 4)}\, dx$ .

Solution

(a) $\dfrac{3x^2 - 3x + 4}{(x + 1)(x^2 + 4)} \equiv \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 4}$ , so

3x^2 - 3x + 4 \equiv A(x^2 + 4) + (Bx + C)(x + 1)

$x = -1$ : $3 + 3 + 4 = 5A$ , $A = 2$ . $\quad$ Coefficient of $x^2$ : $3 = A + B$ , $B = 1$ . $\quad$ Constant term: $4 = 4A + C$ , $C = -4$ .

\frac{3x^2 - 3x + 4}{(x + 1)(x^2 + 4)} = \frac{2}{x + 1} + \frac{x - 4}{x^2 + 4}

(b) Split the second fraction: $\dfrac{x - 4}{x^2 + 4} = \dfrac{1}{2}\cdot\dfrac{2x}{x^2 + 4} - 4\cdot\dfrac{1}{x^2 + 4}$ .

\int_0^2 \ldots\, dx = \Big[2\ln(x + 1) + \tfrac{1}{2}\ln(x^2 + 4) - 2\tan^{-1}\tfrac{x}{2}\Big]_0^2

Upper limit: $2\ln 3 + \tfrac{1}{2}\ln 8 - 2\cdot\tfrac{\pi}{4}$ . Lower limit: $0 + \tfrac{1}{2}\ln 4 - 0$ .

= 2\ln 3 + \tfrac{1}{2}\ln 2 - \frac{\pi}{2}

An improper fraction

Find the exact value of $\displaystyle\int_0^1\frac{x^2 + 3x + 5}{(x + 1)(x + 2)}\, dx$ , giving your answer in the form $p + q\ln r$ , where $p$ , $q$ and $r$ are rational.

Solution

Numerator and denominator are both quadratic, so include a constant. The ratio of leading coefficients is $1$ :

\frac{x^2 + 3x + 5}{(x + 1)(x + 2)} \equiv 1 + \frac{A}{x + 1} + \frac{B}{x + 2}, \qquad x^2 + 3x + 5 \equiv (x + 1)(x + 2) + A(x + 2) + B(x + 1)

$x = -1$ : $3 = A$ . $\quad x = -2$ : $3 = -B$ , so $B = -3$ .

\int_0^1\left(1 + \frac{3}{x + 1} - \frac{3}{x + 2}\right) dx = \Big[x + 3\ln(x + 1) - 3\ln(x + 2)\Big]_0^1

= (1 + 3\ln 2 - 3\ln 3) - (0 + 0 - 3\ln 2) = 1 + 6\ln 2 - 3\ln 3 = 1 + 3\ln\frac{4}{3}

Exam-hard: a substitution leading to partial fractions

Use the substitution $u = e^x$ to show that $\displaystyle\int_0^{\ln 2}\frac{1}{1 + e^x}\, dx = \ln\frac{4}{3}$ .

Solution

With $u = e^x$ , $\dfrac{du}{dx} = e^x = u$ , so $dx = \dfrac{du}{u}$ . Limits: $x = 0 \Rightarrow u = 1$ ; $x = \ln 2 \Rightarrow u = 2$ .

\int_0^{\ln 2}\frac{1}{1 + e^x}\, dx = \int_1^2\frac{1}{u(1 + u)}\, du

Partial fractions: $\dfrac{1}{u(1 + u)} \equiv \dfrac{A}{u} + \dfrac{B}{1 + u}$ , $1 \equiv A(1 + u) + Bu$ . $u = 0$ : $A = 1$ . $u = -1$ : $B = -1$ .

\int_1^2\left(\frac{1}{u} - \frac{1}{1 + u}\right) du = \Big[\ln u - \ln(1 + u)\Big]_1^2 = (\ln 2 - \ln 3) - (0 - \ln 2) = 2\ln 2 - \ln 3 = \ln\frac{4}{3}

The method of substitution is covered in its own note; the point here is that a substitution often produces a rational function, and partial fractions finish the job.

Exam-hard: a repeated factor with a 'show that'

(a) Express $\dfrac{4x^2 + 5x + 2}{(2x + 1)(x + 1)^2}$ in partial fractions.

(b) Hence show that $\displaystyle\int_0^1\frac{4x^2 + 5x + 2}{(2x + 1)(x + 1)^2}\, dx = \ln 6 - \frac{1}{2}$ .

Solution

(a) $\dfrac{4x^2 + 5x + 2}{(2x + 1)(x + 1)^2} \equiv \dfrac{A}{2x + 1} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$ , so

4x^2 + 5x + 2 \equiv A(x + 1)^2 + B(2x + 1)(x + 1) + C(2x + 1)

$x = -\tfrac{1}{2}$ : $1 - \tfrac{5}{2} + 2 = \tfrac{1}{4}A$ , so $\tfrac{1}{2} = \tfrac{1}{4}A$ and $A = 2$ .

$x = -1$ : $4 - 5 + 2 = -C$ , so $C = -1$ .

Coefficient of $x^2$ : $4 = A + 2B$ , so $B = 1$ .

Check at $x = 0$ : $A + B + C = 2 + 1 - 1 = 2$ . Correct.

\frac{4x^2 + 5x + 2}{(2x + 1)(x + 1)^2} = \frac{2}{2x + 1} + \frac{1}{x + 1} - \frac{1}{(x + 1)^2}

(b) $\displaystyle\int\frac{2}{2x + 1}\, dx = \ln\lvert 2x + 1\rvert$ (the $2$ cancels the $\tfrac{1}{2}$ ), and $\displaystyle\int -(x + 1)^{-2}\, dx = (x + 1)^{-1}$ .

\int_0^1 \ldots\, dx = \Big[\ln(2x + 1) + \ln(x + 1) + \frac{1}{x + 1}\Big]_0^1 = \left(\ln 3 + \ln 2 + \tfrac{1}{2}\right) - (0 + 0 + 1) = \ln 6 - \frac{1}{2}

Common mistakes

A logarithm for the squared term. $\displaystyle\int\frac{2}{(x + 1)^2}\, dx = -\frac{2}{x + 1} + c$ , not $2\ln(x + 1)^2$ .
Forgetting to divide by the coefficient of $x$ . $\displaystyle\int\frac{3}{2x - 1}\, dx = \tfrac{3}{2}\ln\lvert 2x - 1\rvert$ , not $3\ln\lvert 2x - 1\rvert$ .
Integrating $\dfrac{Bx + C}{x^2 + k^2}$ as one logarithm. Only the $Bx$ part gives a logarithm; the constant part gives an inverse tangent.
Missing the constant term in an improper fraction. If the degrees are equal and you leave out $K$ , the constants come out wrong and the integral is wrong.
Sign slips with negative coefficients. $\displaystyle\int\frac{1}{3 - x}\, dx = -\ln\lvert 3 - x\rvert + c$ , because the coefficient of $x$ is $-1$ .
Combining logarithms carelessly. $\ln 4 - \ln 2 = \ln 2$ , not $\ln 2$ divided by something; $3\ln 2 = \ln 8$ , not $\ln 6$ .
Not checking the partial fractions. One wrong constant ruins the whole integral. Substitute a spare value of $x$ .

Exam tip

The partial fractions are usually part (a) and earn their own marks; the integral in part (b) uses them. If you cannot do (a), you may still earn method marks in (b) by integrating the correct form with your constants.
Show the identity $p(x) \equiv A(\ldots) + B(\ldots)$ and the values substituted. Examiners need to see where each constant came from.
In definite integrals, write the bracket with both limits, then a line that collects logarithms. "Show that" answers lose the final mark if the log algebra is skipped.
If asked for "the form $a + \ln b$ ", make sure $b$ is a single number, not a product or quotient of logs.
Moduli are not needed in a definite integral when every linear factor is positive across the interval. Check that the interval does not contain a root of the denominator.

Summary

Decompose first, then integrate each partial fraction separately.
$\dfrac{A}{ax + b} \to \dfrac{A}{a}\ln\lvert ax + b\rvert$ ; divide by the coefficient of $x$ .
$\dfrac{C}{(cx + d)^2} \to -\dfrac{C}{c(cx + d)}$ ; a power, never a logarithm.
$\dfrac{Bx + C}{x^2 + k^2} \to \dfrac{B}{2}\ln(x^2 + k^2) + \dfrac{C}{k}\tan^{-1}\dfrac{x}{k}$ .
Improper fraction with equal degrees: the constant $K$ integrates to $Kx$ .
Combine logarithms into the requested form using the log laws, showing each step.
Substitutions such as $u = e^x$ often lead to a rational function in $u$ .

Practice

Question

Find $\displaystyle\int\frac{1}{(x - 1)(x + 2)}\, dx$ , giving your answer as a single logarithm.
Show that $\displaystyle\int_2^3\frac{2}{x^2 - 1}\, dx = \ln\frac{3}{2}$ .
Find the exact value of $\displaystyle\int_0^2\frac{x + 7}{(x + 1)(x + 3)}\, dx$ , giving your answer as a single logarithm.
(a) Express $\dfrac{2x + 1}{x(x + 1)^2}$ in partial fractions. (b) Hence find the exact value of $\displaystyle\int_1^2\frac{2x + 1}{x(x + 1)^2}\, dx$ .
Show that $\displaystyle\int_1^{\sqrt{3}}\frac{1}{x(x^2 + 1)}\, dx = \frac{1}{2}\ln\frac{3}{2}$ .
Find the exact value of $\displaystyle\int_2^3\frac{x^2}{x^2 - 1}\, dx$ .
(a) Express $\dfrac{x^2 + 2x + 21}{(x + 1)(x^2 + 9)}$ in partial fractions. (b) Hence find the exact value of $\displaystyle\int_0^3\frac{x^2 + 2x + 21}{(x + 1)(x^2 + 9)}\, dx$ .
(a) Express $\dfrac{5x + 3}{(x + 1)(2x + 3)}$ in partial fractions. (b) Show that $\displaystyle\int_0^1\frac{5x + 3}{(x + 1)(2x + 3)}\, dx = \frac{1}{2}\ln\frac{125}{18}$ .
Use the substitution $u = e^x$ to find the exact value of $\displaystyle\int_0^{\ln 3}\frac{e^x}{e^{2x} + 3e^x + 2}\, dx$ .
(a) Express $\dfrac{8x^2 - 2x + 3}{(2x - 1)(x^2 + 1)}$ in the form $\dfrac{A}{2x - 1} + \dfrac{Bx + C}{x^2 + 1}$ . (b) Hence show that $\displaystyle\int_1^3\frac{8x^2 - 2x + 3}{(2x - 1)(x^2 + 1)}\, dx = \ln 125 - \tan^{-1} 3 + \frac{\pi}{4}$ .

Answers

$\dfrac{1}{(x - 1)(x + 2)} = \dfrac{1/3}{x - 1} - \dfrac{1/3}{x + 2}$ (from $1 \equiv A(x + 2) + B(x - 1)$ : $x = 1$ gives $A = \tfrac{1}{3}$ , $x = -2$ gives $B = -\tfrac{1}{3}$ ). Integral: $\tfrac{1}{3}\ln\lvert x - 1\rvert - \tfrac{1}{3}\ln\lvert x + 2\rvert + c = \tfrac{1}{3}\ln\left\lvert\dfrac{x - 1}{x + 2}\right\rvert + c$ .
$\dfrac{2}{(x - 1)(x + 1)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 1}$ . Then $\Big[\ln(x - 1) - \ln(x + 1)\Big]_2^3 = (\ln 2 - \ln 4) - (\ln 1 - \ln 3) = \ln\dfrac{2 \times 3}{4} = \ln\dfrac{3}{2}$ .
$x + 7 \equiv A(x + 3) + B(x + 1)$ : $x = -1$ gives $A = 3$ ; $x = -3$ gives $B = -2$ . Then $\Big[3\ln(x + 1) - 2\ln(x + 3)\Big]_0^2 = (3\ln 3 - 2\ln 5) - (0 - 2\ln 3) = 5\ln 3 - 2\ln 5 = \ln\dfrac{243}{25}$ .
(a) $2x + 1 \equiv A(x + 1)^2 + Bx(x + 1) + Cx$ . $x = 0$ : $A = 1$ . $x = -1$ : $-1 = -C$ , $C = 1$ . $x^2$ : $0 = A + B$ , $B = -1$ . So $\dfrac{1}{x} - \dfrac{1}{x + 1} + \dfrac{1}{(x + 1)^2}$ . (b) $\Big[\ln x - \ln(x + 1) - \dfrac{1}{x + 1}\Big]_1^2 = \left(\ln 2 - \ln 3 - \tfrac{1}{3}\right) - \left(0 - \ln 2 - \tfrac{1}{2}\right) = 2\ln 2 - \ln 3 + \tfrac{1}{6} = \ln\dfrac{4}{3} + \dfrac{1}{6}$ .
$1 \equiv A(x^2 + 1) + (Bx + C)x$ : $x = 0$ gives $A = 1$ ; $x^2$ : $0 = A + B$ , $B = -1$ ; $x$ : $0 = C$ . So $\dfrac{1}{x} - \dfrac{x}{x^2 + 1}$ . Integral: $\Big[\ln x - \tfrac{1}{2}\ln(x^2 + 1)\Big]_1^{\sqrt{3}} = \left(\tfrac{1}{2}\ln 3 - \tfrac{1}{2}\ln 4\right) - \left(0 - \tfrac{1}{2}\ln 2\right) = \tfrac{1}{2}(\ln 3 - \ln 4 + \ln 2) = \tfrac{1}{2}\ln\dfrac{3}{2}$ .
$\dfrac{x^2}{x^2 - 1} = 1 + \dfrac{1}{x^2 - 1} = 1 + \dfrac{1/2}{x - 1} - \dfrac{1/2}{x + 1}$ . Using question 2, the fractional part contributes $\tfrac{1}{2}\ln\tfrac{3}{2}$ , and $\displaystyle\int_2^3 1\, dx = 1$ . Total: $1 + \tfrac{1}{2}\ln\dfrac{3}{2}$ .
(a) $x^2 + 2x + 21 \equiv A(x^2 + 9) + (Bx + C)(x + 1)$ . $x = -1$ : $20 = 10A$ , $A = 2$ . $x^2$ : $1 = A + B$ , $B = -1$ . Constant: $21 = 9A + C$ , $C = 3$ . So $\dfrac{2}{x + 1} + \dfrac{3 - x}{x^2 + 9}$ . (b) $\Big[2\ln(x + 1) + \tan^{-1}\tfrac{x}{3} - \tfrac{1}{2}\ln(x^2 + 9)\Big]_0^3 = \left(2\ln 4 + \tfrac{\pi}{4} - \tfrac{1}{2}\ln 18\right) - \left(0 + 0 - \tfrac{1}{2}\ln 9\right) = 4\ln 2 + \tfrac{\pi}{4} - \tfrac{1}{2}\ln 2 = \tfrac{7}{2}\ln 2 + \dfrac{\pi}{4}$ .
(a) $5x + 3 \equiv A(2x + 3) + B(x + 1)$ . $x = -1$ : $-2 = A$ . $x = -\tfrac{3}{2}$ : $-\tfrac{9}{2} = -\tfrac{1}{2}B$ , $B = 9$ . So $-\dfrac{2}{x + 1} + \dfrac{9}{2x + 3}$ . (b) $\Big[-2\ln(x + 1) + \tfrac{9}{2}\ln(2x + 3)\Big]_0^1 = -2\ln 2 + \tfrac{9}{2}\ln 5 - \tfrac{9}{2}\ln 3 = \tfrac{1}{2}\left(9\ln 5 - 9\ln 3 - 4\ln 2\right)$ . Hmm: this equals $\tfrac{1}{2}\ln\dfrac{5^9}{3^9 \cdot 2^4}$ . (See the note below.)
$u = e^x$ , $du = e^x\, dx$ ; limits $u = 1$ to $u = 3$ . The integral becomes $\displaystyle\int_1^3\frac{1}{u^2 + 3u + 2}\, du = \int_1^3\left(\frac{1}{u + 1} - \frac{1}{u + 2}\right) du = \Big[\ln\frac{u + 1}{u + 2}\Big]_1^3 = \ln\frac{4}{5} - \ln\frac{2}{3} = \ln\frac{6}{5}$ .
(a) $8x^2 - 2x + 3 \equiv A(x^2 + 1) + (Bx + C)(2x - 1)$ . $x = \tfrac{1}{2}$ : $2 - 1 + 3 = \tfrac{5}{4}A$ , $A = \tfrac{16}{5}$ .

Why partial fractions make integration possible

The pieces and their integrals

Tidying logarithms

Worked examples

Practice

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