Factor forms and required numerators
Proper versus improper rule: partial-fraction templates apply when degree P < degree Q. If degree P ≥ degree Q, first divide to write P/Q = S + R/Q, then decompose the proper remainder R/Q.
| Factor case | Denominator example | Required numerator | Complete template contribution |
|---|---|---|---|
| Distinct linear | (x − a)(x − b) | One constant per factor | A/(x − a) + B/(x − b) |
| Repeated linear | (x − a)m | A constant over every power | A1/(x − a) + A2/(x − a)2 + … + Am/(x − a)m |
| Irreducible quadratic | x2 + bx + c | One degree-lower linear numerator | (Ax + B)/(x2 + bx + c) |
| Repeated quadratic | q(x)m | A linear numerator over every power | (A1x+B1)/q(x) + … + (Amx+Bm)/q(x)m |
Fully worked partial fraction examples
Distinct linear factors
Expression and factorization: (5x + 3)/(x2 − 1), where x2 − 1 = (x − 1)(x + 1).
Template: A/(x − 1) + B/(x + 1).
Coefficient equations: 5x + 3 = A(x + 1) + B(x − 1) = (A+B)x + (A−B), so A+B=5 and A−B=3.
Constants and result: A=4, B=1; therefore 4/(x−1) + 1/(x+1).
Verification: 4(x+1)+(x−1)=5x+3.
Repeated linear factor
Expression and factorization: (2x2+3x+4)/[x(x+1)2].
Template: A/x + B/(x+1) + C/(x+1)2.
Coefficient equations: 2x2+3x+4=A(x+1)2+Bx(x+1)+Cx, giving A+B=2, 2A+B+C=3, A=4.
Constants and result: A=4, B=−2, C=−3; therefore 4/x − 2/(x+1) − 3/(x+1)2.
Verification: recombination expands to 2x2+3x+4.
Irreducible quadratic factor
Expression and factorization: (3x2+5x+7)/[(x−1)(x2+4)].
Template: A/(x−1) + (Bx+C)/(x2+4).
Coefficient equations: A+B=3, C−B=5, 4A−C=7.
Constants and result: A=3, B=0, C=5; therefore 3/(x−1) + 5/(x2+4).
Verification: 3(x2+4)+5(x−1)=3x2+5x+7.
Improper fraction
Expression and division: (x3+2x2+3)/(x2−1) = x+2 + (x+5)/(x2−1).
Factorization and template: x2−1=(x−1)(x+1), so use x+2 + A/(x−1)+B/(x+1).
Coefficient equations: x+5=A(x+1)+B(x−1), giving A+B=1, A−B=5.
Constants and result: A=3, B=−2; therefore x+2 + 3/(x−1) − 2/(x+1).
Verification: the quotient times x2−1 plus x+5 is x3+2x2+3.
When partial fractions are useful
Decomposition turns a complicated rational function into standard pieces used in integration, inverse Laplace transforms, differential equations, control-system calculations, and algebraic simplification.
Short integration example: if F(x)=3/(x−2)+1/(x+1), then ∫F(x)dx=3 ln|x−2|+ln|x+1|+C. In general, A/(x−a) integrates to A ln|x−a|. A linear numerator over an irreducible quadratic can split into a derivative part, which produces a logarithm, and a remaining constant part, which can produce an arctangent.
Accuracy, verification, and limits
Exact mode: parsing, decimal-to-fraction conversion, polynomial GCD, long division, coefficient elimination, and verification use reduced BigInt rational numbers. The identity check recombines every returned term over the common denominator and compares every numerator coefficient exactly.
Approximate mode: roots and coefficients are computed numerically and rounded for display. It is intended for suitable square-free denominators with irrational real roots or unresolved cubic and higher factors; repeated numerical roots are rejected because rounding can make their multiplicity unreliable.
The practical degree limit is 12. Exact mode requires the reduced denominator to factor over the rationals into linear and quadratic factors. Approximate mode may still reject ill-conditioned or unresolved roots and tells you to enter a known factorization. Original denominator restrictions remain after cancellation.
| Representative test | Expected result | Evidence |
|---|---|---|
| Distinct linear | (5x+3)/(x²−1) | 4/(x−1)+1/(x+1), exact identity |
| Repeated linear | (2x²+3x+4)/[x(x+1)²] | 4/x−2/(x+1)−3/(x+1)², exact identity |
| Quadratic | (3x²+5x+7)/[(x−1)(x²+4)] | 3/(x−1)+5/(x²+4), exact identity |
| Improper | (x³+2x²+3)/(x²−1) | x+2+3/(x−1)−2/(x+1), exact identity |
References: OpenStax Calculus, §3.4 Partial Fractions and Mathematics LibreTexts, Partial Fractions.
Last reviewed: 31 August 2026. Report an incorrect result and include the expression and displayed steps.
Partial fractions FAQs
When can partial fractions be used?
Use partial fractions for a rational function P(x)/Q(x) after Q(x) is factored. If degree P is at least degree Q, divide first so the remaining fraction is proper.
What are the four denominator factor cases?
The four cases are distinct linear factors, repeated linear factors, irreducible quadratic factors, and repeated irreducible quadratic factors. Repeated factors need one term for every power.
How are the coefficients found?
Clear the denominators, expand every product, collect equal powers of the variable, and solve the resulting simultaneous linear equations.
What is the cover-up method?
For distinct linear factors, multiply by one factor and substitute its root to isolate that term's constant. Repeated or quadratic factors generally still require substitution plus coefficient comparison.
Why must an improper fraction be divided first?
Partial-fraction templates apply to a proper remainder. Polynomial division writes an improper rational function as a polynomial quotient plus R(x)/Q(x), where degree R is less than degree Q.
Can decomposition be used before integration or inverse Laplace transforms?
Yes. It turns one complicated rational expression into standard terms that are easier to integrate or invert term by term.
Why was my denominator rejected?
Exact mode needs a degree-12-or-lower denominator that factors over the rationals into linear and quadratic factors. Try Approximate real factors for a suitable unresolved cubic or higher factor, or enter a known factorization.
Does decomposition change the domain?
No. Every zero of the original denominator remains excluded, including a zero belonging to a factor that cancels during simplification.