Integration by Partial Fractions

Quick answer

A rational function whose denominator factors can be split into simpler fractions, each of which integrates to a logarithm: 1/((x − 1)(x + 2)) = (1/3)/(x − 1) − (1/3)/(x + 2). Write one term A/(x − r) for each distinct linear factor, find the constants by clearing denominators and substituting convenient values of x, then integrate term by term. Divide first if the numerator's degree is not lower than the denominator's; a repeated factor (x − r)² needs the two terms A/(x − r) and B/(x − r)².

What you'll learn

Decompose a rational function with distinct linear factors
Find the constants by substituting convenient values
Integrate the decomposition term by term
Handle top-heavy fractions and repeated factors

Adding fractions, run backward

Adding two fractions over a common denominator merges them:

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

Integrating the left side is immediate: two logarithms. Integrating the right side is not. Partial fractions runs the addition backward, splitting a fraction into pieces that each integrate to a logarithm.

The method

For a fraction $\tfrac{P(x)}{Q(x)}$ whose denominator factors into distinct linear factors:

If the degree of $P$ is not less than the degree of $Q$ , divide first.
Factor $Q$ completely.
Write one term $\tfrac{A}{x - r}$ for each factor $x - r$ .
Multiply through by $Q$ and find the constants.
Integrate: each $\tfrac{A}{x - r}$ gives $A\ln\lvert x - r \rvert$ .

Why substituting the roots finds the constants

Take $\tfrac{1}{(x - 1)(x + 2)} = \tfrac{A}{x - 1} + \tfrac{B}{x + 2}$ . Multiplying by $(x - 1)(x + 2)$ gives

1 = A(x + 2) + B(x - 1)

This is an identity: it holds for every $x$ , not only for one. So choose the values of $x$ that make terms vanish. At $x = 1$ the $B$ term drops out and $1 = 3A$ . At $x = -2$ the $A$ term drops out and $1 = -3B$ . The decomposition is an identity, so substitute the roots, and each one reveals a constant.

Worked examples

Example 2 — Three linear factors

Find $\displaystyle\int \frac{x + 5}{x^3 - x}\,dx$ .

$x^3 - x = x(x - 1)(x + 1)$ , so write $\tfrac{x + 5}{x^3 - x} = \tfrac{A}{x} + \tfrac{B}{x - 1} + \tfrac{C}{x + 1}$ and clear denominators:

x + 5 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)

At $x = 0$ : $5 = -A$ . At $x = 1$ : $6 = 2B$ . At $x = -1$ : $4 = 2C$ . So $A = -5$ , $B = 3$ , $C = 2$ , and

\int \frac{x + 5}{x^3 - x}\,dx = -5\ln\lvert x \rvert + 3\ln\lvert x - 1 \rvert + 2\ln\lvert x + 1 \rvert + C

Example 4 — A repeated factor

Find $\displaystyle\int \frac{dx}{x(x - 1)^2}$ .

A squared factor needs a term for each power: $\tfrac{1}{x(x - 1)^2} = \tfrac{A}{x} + \tfrac{B}{x - 1} + \tfrac{C}{(x - 1)^2}$ , so

1 = A(x - 1)^2 + Bx(x - 1) + Cx

At $x = 0$ , $A = 1$ . At $x = 1$ , $C = 1$ . No root isolates $B$ , so compare the $x^2$ terms: $0 = A + B$ , so $B = -1$ . Since $\int (x - 1)^{-2}\,dx = -(x - 1)^{-1}$ ,

\int \frac{dx}{x(x - 1)^2} = \ln\lvert x \rvert - \ln\lvert x - 1 \rvert - \frac{1}{x - 1} + C

Common mistakes

Practice problems

Find $\displaystyle\int \frac{dx}{(x - 2)(x + 3)}$ .

Answer

$\tfrac{1}{5}\ln\left\lvert \tfrac{x - 2}{x + 3} \right\rvert + C$

Full solution

$1 = A(x + 3) + B(x - 2)$ . At $x = 2$ , $A = \tfrac{1}{5}$ ; at $x = -3$ , $B = -\tfrac{1}{5}$ .
Find $\displaystyle\int \frac{5}{x^2 + x - 6}\,dx$ .

Answer

$\ln\left\lvert \tfrac{x - 2}{x + 3} \right\rvert + C$

Full solution

$x^2 + x - 6 = (x - 2)(x + 3)$ , and $5 = A(x + 3) + B(x - 2)$ gives $A = 1$ and $B = -1$ .
Find $\displaystyle\int \frac{3x + 1}{x^2 - 1}\,dx$ .

Answer

$2\ln\lvert x - 1 \rvert + \ln\lvert x + 1 \rvert + C$

Full solution

$3x + 1 = A(x + 1) + B(x - 1)$ . At $x = 1$ , $4 = 2A$ ; at $x = -1$ , $-2 = -2B$ .
Find $\displaystyle\int \frac{dx}{x^2 - 4}$ .

Answer

$\tfrac{1}{4}\ln\left\lvert \tfrac{x - 2}{x + 2} \right\rvert + C$

Full solution

$1 = A(x + 2) + B(x - 2)$ gives $A = \tfrac{1}{4}$ and $B = -\tfrac{1}{4}$ .
Find $\displaystyle\int \frac{x + 1}{x^2 - 3x}\,dx$ .

Answer

$-\tfrac{1}{3}\ln\lvert x \rvert + \tfrac{4}{3}\ln\lvert x - 3 \rvert + C$

Full solution

$x^2 - 3x = x(x - 3)$ , and $x + 1 = A(x - 3) + Bx$ . At $x = 0$ , $1 = -3A$ ; at $x = 3$ , $4 = 3B$ .
Find $\displaystyle\int \frac{dx}{2x^2 + x - 1}$ .

Answer

$\tfrac{1}{3}\ln\lvert 2x - 1 \rvert - \tfrac{1}{3}\ln\lvert x + 1 \rvert + C$

Full solution

$2x^2 + x - 1 = (2x - 1)(x + 1)$ , and $1 = A(x + 1) + B(2x - 1)$ . At $x = \tfrac{1}{2}$ , $A = \tfrac{2}{3}$ ; at $x = -1$ , $B = -\tfrac{1}{3}$ . Then $\int \tfrac{2/3}{2x - 1}\,dx = \tfrac{1}{3}\ln\lvert 2x - 1 \rvert$ .
Evaluate $\displaystyle\int_0^1 \frac{dx}{x^2 + 3x + 2}$ .

Answer

$\ln\tfrac{4}{3} \approx 0.288$

Full solution

$\tfrac{1}{(x + 1)(x + 2)} = \tfrac{1}{x + 1} - \tfrac{1}{x + 2}$ , so the integral is $\big[\ln\tfrac{x + 1}{x + 2}\big]_0^1 = \ln\tfrac{2}{3} - \ln\tfrac{1}{2} = \ln\tfrac{4}{3}$ .
Find $\displaystyle\int \frac{x^2 + 1}{x^2 - 1}\,dx$ .

Answer

$x + \ln\left\lvert \tfrac{x - 1}{x + 1} \right\rvert + C$

Full solution

Divide: $\tfrac{x^2 + 1}{x^2 - 1} = 1 + \tfrac{2}{x^2 - 1}$ , and $\tfrac{2}{x^2 - 1} = \tfrac{1}{x - 1} - \tfrac{1}{x + 1}$ .
Find $\displaystyle\int \frac{dx}{x^2(x + 1)}$ .

Answer

$-\ln\lvert x \rvert - \tfrac{1}{x} + \ln\lvert x + 1 \rvert + C$

Full solution

$1 = Ax(x + 1) + B(x + 1) + Cx^2$ . At $x = 0$ , $B = 1$ ; at $x = -1$ , $C = 1$ ; the $x^2$ terms give $0 = A + C$ , so $A = -1$ . Then $\int x^{-2}\,dx = -\tfrac{1}{x}$ .
A student writes $\tfrac{1}{x^2 - 9} = \tfrac{1}{x^2} - \tfrac{1}{9}$ and integrates to $-\tfrac{1}{x} - \tfrac{x}{9} + C$ . What went wrong?

Hint

Test the student’s first step with $x = 4$ .

Answer

A fraction does not split over a difference in its denominator. The integral is $\tfrac{1}{6}\ln\left\lvert \tfrac{x - 3}{x + 3} \right\rvert + C$ .

Full solution

At $x = 4$ , $\tfrac{1}{16 - 9} = \tfrac{1}{7}$ , but $\tfrac{1}{16} - \tfrac{1}{9}$ is negative.

Factor instead: $x^2 - 9 = (x - 3)(x + 3)$ and $1 = A(x + 3) + B(x - 3)$ give $A = \tfrac{1}{6}$ and $B = -\tfrac{1}{6}$ .

Frequently asked questions

What is partial fraction decomposition?

Rewriting a fraction of polynomials as a sum of simpler fractions, one for each factor of the denominator. It is adding fractions run backward.

How do I find the constants A and B?

Multiply both sides by the denominator. The result holds for every x, so substitute each root of the denominator: each substitution makes all but one term vanish.

What if the numerator's degree is too high?

Divide first. Partial fractions apply only to the proper fraction left after long division.

What do I do with a repeated factor like (x − 1)²?

Include a term for each power: A/(x − 1) + B/(x − 1)². The second term integrates to −B/(x − 1).

What about a factor like x² + 1 that does not factor?

Give it a term (Bx + C)/(x² + 1), which integrates to a logarithm and an arctangent after completing the square.

What to learn next

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