Partial fractions, explained with examples

By The STEM-Tutor team · Updated October 2026

Partial fractions split one complicated fraction into a sum of simpler ones. It is the reverse of adding fractions, and it makes difficult integrals and expansions straightforward.

The idea

Adding fractions is familiar: 2/(x + 1) + 3/(x − 2) = (5x − 1)/((x + 1)(x − 2)). Partial fractions go the other way. Given (5x − 1)/((x + 1)(x − 2)), we want to find the 2 and the 3. The split form is much easier to integrate or expand, because each simple fraction can be handled on its own.

The three forms you need

• Distinct linear factors: (px + q)/((x − a)(x − b)) = A/(x − a) + B/(x − b).
• A repeated linear factor: f(x)/((x − a)(x − b)²) = A/(x − a) + B/(x − b) + C/(x − b)². The repeated factor needs one fraction for each power.
• Improper fractions: if the degree of the top is equal to or greater than the degree of the bottom, divide first so you have a polynomial plus a proper fraction.

Method 1: substitution

1. Write the fraction in the correct split form with unknown constants A, B, ….
2. Multiply both sides by the denominator to clear fractions.
3. Substitute values of x that make factors zero. Each one removes all but one unknown.
4. If needed, compare coefficients or substitute another convenient value such as x = 0.

Worked examples

Example 1: two linear factors

Express (5x − 1)/((x + 1)(x − 2)) in partial fractions.

• Write (5x − 1)/((x + 1)(x − 2)) ≡ A/(x + 1) + B/(x − 2).
• Multiply through: 5x − 1 ≡ A(x − 2) + B(x + 1).
• x = 2: 9 = 3B, so B = 3. x = −1: −6 = −3A, so A = 2.
• 2/(x + 1) + 3/(x − 2).

Example 2: three linear factors

Express (x² + 3x − 2)/(x(x − 1)(x + 2)) in partial fractions.

• x² + 3x − 2 ≡ A(x − 1)(x + 2) + Bx(x + 2) + Cx(x − 1).
• x = 0: −2 = −2A, so A = 1. x = 1: 2 = 3B, so B = 2/3. x = −2: −4 = 6C, so C = −2/3.
• 1/x + 2/(3(x − 1)) − 2/(3(x + 2)).

Example 3: a repeated factor

Express (3x + 1)/((x − 1)(x + 1)²) in partial fractions.

• 3x + 1 ≡ A(x + 1)² + B(x − 1)(x + 1) + C(x − 1).
• x = 1: 4 = 4A, so A = 1. x = −1: −2 = −2C, so C = 1.
• Compare x² coefficients: 0 = A + B, so B = −1.
• 1/(x − 1) − 1/(x + 1) + 1/(x + 1)².

Example 4: an improper fraction

Express (x² + 2)/((x + 1)(x − 1)) in partial fractions.

• Top and bottom are both degree 2, so divide: (x² + 2)/(x² − 1) = 1 + 3/(x² − 1).
• 3/((x + 1)(x − 1)) = A/(x + 1) + B/(x − 1): x = 1 gives B = 3/2; x = −1 gives A = −3/2.
• 1 − 3/(2(x + 1)) + 3/(2(x − 1)).

Method 2: the cover-up rule

For distinct linear factors, there is a quick shortcut. To find the constant over (x − a), cover up that factor in the original fraction and substitute x = a into what remains. In Example 1, to find B, cover (x − 2) and put x = 2 into (5x − 1)/(x + 1): 9/3 = 3. To find A, cover (x + 1) and put x = −1 into (5x − 1)/(x − 2): −6/−3 = 2. It is fast, but in an exam, show enough working that the examiner can follow it.

Checking your answer

A quick check is to substitute a simple value, such as x = 0 or x = 3, into both the original fraction and your partial fractions. If they give the same number, your constants are almost certainly right. For Example 1 at x = 0: the original gives −1/−2 = 1/2, and 2/1 + 3/(−2) = 2 − 1.5 = 1/2. It takes seconds and catches most sign slips.

Using partial fractions in integration

Each simple fraction integrates to a logarithm or a power, so partial fractions make many integrals routine.

Example 5: integrating

Find ∫ (5x − 1)/((x + 1)(x − 2)) dx.

• From Example 1, the integrand is 2/(x + 1) + 3/(x − 2).
• = 2 ln|x + 1| + 3 ln|x − 2| + C.

Example 6: a definite integral

Evaluate ∫₂³ 1/(x(x − 1)) dx, giving your answer as a single logarithm.

• 1/(x(x − 1)) = 1/(x − 1) − 1/x.
• [ln|x − 1| − ln|x|] from 2 to 3 = (ln 2 − ln 3) − (0 − ln 2) = 2 ln 2 − ln 3.
• = ln(4/3).

Using partial fractions with binomial expansion

To expand a fraction such as (5x − 1)/((x + 1)(x − 2)) as a series, split it first, then expand each part. Rewrite 3/(x − 2) as −(3/2)(1 − x/2)⁻¹ so the bracket starts with 1, then apply the binomial series. The overall expansion is valid only where both parts are valid: here |x| < 1 from the first part and |x| < 2 from the second, so |x| < 1.

Common mistakes

• Using only one fraction for a repeated factor, instead of one for each power.
• Trying partial fractions on an improper fraction without dividing first.
• Sign errors when substituting negative values of x.
• Forgetting the modulus signs in ln|x + 1| after integrating.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Are partial fractions in A level Maths?

Yes, in the second year of A level Maths, usually with linear and repeated linear factors. Quadratic factors that do not factorise are in Further Maths.

Are partial fractions on AP Calculus?

Partial fractions with distinct linear factors are part of AP Calculus BC integration. AB does not include them.

Key takeaways

• Partial fractions reverse the addition of fractions.
• Use one fraction per linear factor, and one per power for repeated factors.
• Divide first if the fraction is improper.
• Substitute values that make factors zero to find constants quickly.
• Split first, then integrate or expand each part.

Related guides

• Binomial expansion
• A level Maths integration
• Integration by parts