Why use partial fractions if I can just integrate the original fraction?

Most rational functions cannot be integrated directly. Partial fractions transforms a complicated fraction into simple terms like \frac{A}{x-a}, each of which integrates easily using \int \frac{A}{x-a} dx = A \ln|x-a| + C. Without decomposing, you have no straightforward integration path.

What do I do if the roots of the denominator are repeated, like (x-1)^2 instead of (x-1)(x-2)?

Repeated factors require a different decomposition form: \frac{A}{x-1} + \frac{B}{(x-1)^2}. You must include all powers of the repeated factor, from the first power up to the highest power appearing in the denominator.

Can I use partial fractions if the numerator's degree is larger than the denominator's degree?

Not directly. Use polynomial long division to split the fraction into a polynomial part and a proper rational part (where numerator degree is less than denominator degree). Then apply partial fractions only to the remainder and integrate or analyze both pieces separately.

What is the difference between partial fractions and other integration techniques like Integration by Parts or Trigonometric Substitution?

Partial fractions is specific to rational functions (ratios of polynomials) with factorable denominators. For irreducible quadratic terms in the denominator, use Trigonometric Substitution. For products of different function types, use Integration by Parts. Each technique targets a different function form.

Partial Fractions

Partial Fractions breaks a rational function into a sum of simpler fractions, each of which is easier to integrate or analyze.

\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

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What each symbol means

What Partial Fractions takes

Symbol	Meaning
$P$	The numerator polynomial; confusing it with the full fraction's value obscures the structure of the decomposition.
$A$	The constant numerator of the first decomposed fraction; writing it as a polynomial instead breaks the method.
$B$	The constant numerator of the second decomposed fraction; treating it as a variable rather than a constant defeats the purpose.
$a$	The first root of the denominator (where $x-a = 0$ ); misidentifying it prevents correct substitution when solving for $A$ and $B$ .
$b$	The second root of the denominator (where $x-b = 0$ ); it must be distinct from $a$ , and swapping them produces wrong results.
$x$	The variable in the rational function; treating it as a constant makes it impossible to solve for $A$ and $B$ by substitution.

When to use it

When you need to integrate a rational function or decompose it into simpler components for analysis.

Level

Usually taught in: Calculus II

Worked examples

1. Decompose a fraction with a constant numerator

Problem

Decompose

\frac{5}{(x-1)(x-2)}

into partial fractions.

$\frac{5}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$
Set up the partial fraction form by expressing the single fraction as a sum of fractions with unknown constants $A$ and $B$ .
$5 = A(x-2) + B(x-1)$
Multiply both sides by the common denominator $(x-1)(x-2)$ to clear all fractions.
$x = 1: \quad 5 = A(1-2) + B(0) = -A, \quad A = -5$
Substitute $x = 1$ to eliminate the $B$ term and solve for $A$ : we get $5 = -A$ , so $A = -5$ .
$x = 2: \quad 5 = A(0) + B(2-1) = B, \quad B = 5$
Substitute $x = 2$ to eliminate the $A$ term and solve for $B$ : we get $5 = B$ , so $B = 5$ .

Answer: $\frac{5}{(x-1)(x-2)} = \frac{-5}{x-1} + \frac{5}{x-2}$

We set up the form with unknown constants $A$ and $B$ , then multiply to clear denominators. Substituting the roots $x=1$ and $x=2$ eliminates one term each time, leaving a simple equation for the unknown. This method is quick and reliable for non-repeated linear factors.

2. Decompose a fraction with a linear numerator

Problem

Decompose

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

into partial fractions.

$\frac{3x-1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$
Set up the partial fraction form with unknown constants $A$ and $B$ .
$3x - 1 = A(x-2) + B(x+1)$
Multiply both sides by $(x+1)(x-2)$ to eliminate the denominators.
$x = -1: \quad 3(-1) - 1 = A(-3) + B(0) = -3A, \quad -4 = -3A, \quad A = \frac{4}{3}$
Substitute $x = -1$ to find $A$ : we get $-4 = -3A$ , so $A = \frac{4}{3}$ .
$x = 2: \quad 3(2) - 1 = A(0) + B(3) = 3B, \quad 5 = 3B, \quad B = \frac{5}{3}$
Substitute $x = 2$ to find $B$ : we get $5 = 3B$ , so $B = \frac{5}{3}$ .

Answer: $\frac{3x-1}{(x+1)(x-2)} = \frac{4/3}{x+1} + \frac{5/3}{x-2}$

Even when the numerator is not constant, we apply the same substitution method. The numerator $3x-1$ is a first-degree polynomial, which fits the framework where the numerator degree is strictly less than the denominator degree. The fractional constants $A = \frac{4}{3}$ and $B = \frac{5}{3}$ are normal and expected; they show that partial fractions works beyond just integer solutions.

3. Decompose a rate function in a cyclist's training scenario

Problem

A cyclist rides over a hill. The acceleration at different points on the route (in m/s²) is given by

\frac{70}{(t-1)(t-5)}

, where

t

is time in minutes from the base of the hill. Decompose this into partial fractions.

$\frac{70}{(t-1)(t-5)} = \frac{A}{t-1} + \frac{B}{t-5}$
Set up the partial fraction form with unknown constants $A$ and $B$ .
$70 = A(t-5) + B(t-1)$
Multiply both sides by $(t-1)(t-5)$ to clear the denominators.
$t = 1: \quad 70 = A(1-5) + B(0) = -4A, \quad A = -\frac{70}{4} = -\frac{35}{2}$
Substitute $t = 1$ to solve for $A$ : we get $70 = -4A$ , so $A = -\frac{35}{2}$ .
$t = 5: \quad 70 = A(0) + B(5-1) = 4B, \quad B = \frac{70}{4} = \frac{35}{2}$
Substitute $t = 5$ to solve for $B$ : we get $70 = 4B$ , so $B = \frac{35}{2}$ .

Answer: $\frac{70}{(t-1)(t-5)} = \frac{-35/2}{t-1} + \frac{35/2}{t-5}$

In applied contexts, partial fractions decompose a complex rate into interpretable components. Here, the negative and positive terms represent different physical effects contributing to the cyclist's acceleration profile up and down the hill. Once decomposed, each term integrates easily using $\int \frac{c}{t-a} dt = c \ln|t-a| + C$ , making it straightforward to calculate total displacement over any time interval.

Common mistakes

Where Partial Fractions usually goes wrong

The mistake	Why it is wrong	The fix
$\frac{5}{(x-1)(x-2)} = \frac{Ax}{x-1} + \frac{Bx}{x-2}$	The numerators must be constants, not polynomials or multiples of $x$ , when decomposing into simple linear factors.	Use $\frac{A}{x-1} + \frac{B}{x-2}$ with $A$ and $B$ as constants only.
Starting with $\frac{5}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$ and immediately substituting $x=1$ to get $\frac{5}{0} = \frac{A}{0} + \frac{B}{-1}$	Substituting into the original unsimplified equation produces undefined expressions; you must clear denominators first.	Multiply both sides by $(x-1)(x-2)$ first to get $5 = A(x-2) + B(x-1)$ , then substitute.
$\frac{7}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{x-1}$	Repeated factors require different numerator forms with increasing powers; writing two identical terms does not give enough flexibility to solve for the constants.	For $(x-1)^2$ , write $\frac{A}{x-1} + \frac{B}{(x-1)^2}$ , not two fractions with the same denominator.

Tips and when to use something else

Verify your answer by combining the decomposed fractions back together: add them using a common denominator and check that you recover the original fraction.
If the numerator's degree is greater than or equal to the denominator's degree, first perform polynomial long division; the remainder becomes the numerator for partial fractions.
The power of partial fractions is that each term $\frac{A}{x-a}$ integrates to $A \ln|x-a| + C$ , transforming a difficult rational integral into a sum of logarithms.
If the denominator contains an irreducible quadratic factor like $x^2+1$ or $x^2+x+1$ , use Trigonometric Substitution instead of partial fractions.

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.Integration by PartsIntegration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.

Frequently asked questions

Why use partial fractions if I can just integrate the original fraction?: Most rational functions cannot be integrated directly. Partial fractions transforms a complicated fraction into simple terms like $\frac{A}{x-a}$ , each of which integrates easily using $\int \frac{A}{x-a} dx = A \ln|x-a| + C$ . Without decomposing, you have no straightforward integration path.
What do I do if the roots of the denominator are repeated, like $(x-1)^2$ instead of $(x-1)(x-2)$ ?: Repeated factors require a different decomposition form: $\frac{A}{x-1} + \frac{B}{(x-1)^2}$ . You must include all powers of the repeated factor, from the first power up to the highest power appearing in the denominator.
Can I use partial fractions if the numerator's degree is larger than the denominator's degree?: Not directly. Use polynomial long division to split the fraction into a polynomial part and a proper rational part (where numerator degree is less than denominator degree). Then apply partial fractions only to the remainder and integrate or analyze both pieces separately.
What is the difference between partial fractions and other integration techniques like Integration by Parts or Trigonometric Substitution?: Partial fractions is specific to rational functions (ratios of polynomials) with factorable denominators. For irreducible quadratic terms in the denominator, use Trigonometric Substitution. For products of different function types, use Integration by Parts. Each technique targets a different function form.

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Reviewed 2026-09-18