This partial fractions calculator decomposes a proper rational function (ax + b) / (x − r₁)(x − r₂) into A/(x − r₁) + B/(x − r₂) — and handles the repeated-factor case (ax + b)/(x − r)² as A/(x − r) + B/(x − r)². Partial fractions turn one hard fraction into several easy ones: the move that makes rational functions integrable and Laplace transforms invertible.

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Why One Fraction Becomes Two: The Cover-Up

Partial fractions rest on a tidy theorem: any proper rational function with a factorable denominator splits into a sum of simplest-possible fractions, one per factor, with constant numerators (for linear factors). The Heaviside cover-up method solves the constants almost by insult: to find A in (ax + b)/((x − r₁)(x − r₂)) = A/(x − r₁) + B/(x − r₂), multiply through by (x − r₁) — “covering up” that factor — and set x = r₁; the B-term dies and A falls out as (a·r₁ + b)/(r₁ − r₂). Repeat for B. Behind the trick sits plain algebra (it’s evaluating an identity of polynomials at a root), and the built-in audit is equally plain: A + B must return the numerator’s x-coefficient, and the constant terms must reconcile — this calculator prints both checks with every answer, because a decomposition that doesn’t recombine is just a rumor.

The repeated-root case changes the template: (x − r)² demands two fractions, A/(x − r) and B/(x − r)² — one per power. Skipping the lower-power term is the canonical error (the system then has no solution, which students experience as “the equations contradict”). B is still pure cover-up (evaluate at r); A comes from comparing the x-coefficients (or differentiating the cleared identity, Heaviside’s own extension). Our worked examples run both templates with the checks visible.

Worked Examples (Verified)

  • (5x − 4)/((x − 1)(x + 2)) (defaults; note x + 2 = (x − (−2)), so r₂ = −2): A = (5 − 4)/(1 − (−2)) = 1/3; B = (−10 − 4)/(−2 − 1) = 14/3. Audit: A + B = 15/3 = 5 ✔ (the x-coefficient); constants: A·(−2)… via −A·r₂ − B·r₁ = 2/3 − 14/3 = −4 ✔ (the numerator’s constant). Result: (1/3)/(x − 1) + (14/3)/(x + 2), integrating to (1/3) ln|x − 1| + (14/3) ln|x + 2| + C.
  • Repeated: (3x + 5)/(x − 2)² → B = 3(2) + 5 = 11, A = 3 → 3/(x − 2) + 11/(x − 2)². Recombine mentally at x = 0: LHS 5/4; RHS −3/2 + 11/4 = 5/4 ✔.
  • Why it matters in one line: ∫(5x − 4)/((x − 1)(x + 2)) dx looks like a research project and decomposes into two logarithm integrals a first-year student finishes in a minute — the decomposition IS the method.

Partial-Fraction Discipline

  • Proper first: if the numerator’s degree ≥ the denominator’s, divide first (Polynomial Division Calculator) — the theorem’s simple form covers the proper remainder only.
  • Factor completely before templating: irreducible quadratics in the denominator take (Ax + B)/(quadratic) terms, not constants — a third template this page’s FAQ explains even though the widgets here cover the linear cases.
  • Every distinct factor gets a term; every repeated factor gets a staircase of terms up to its multiplicity. Missing rungs = unsolvable systems.
  • Always recombine-check at a convenient x (x = 0 is the usual witness) before integrating; the calculator prints the coefficient checks for exactly this habit.
  • Sign discipline with (x + 2)-style factors: r₂ = −2, and the decomposed term is /(x + 2) — display forms and r-values must agree, or the cover-up evaluates at the wrong point.

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Frequently Asked Questions

How do you decompose a partial fraction?

Factor the denominator, write one fraction per factor with unknown constant numerators, clear denominators, and solve — fastest by the cover-up: for (5x − 4)/((x − 1)(x + 2)), A = (5·1 − 4)/(1 − (−2)) = 1/3 and B = (5·(−2) − 4)/(−2 − 1) = 14/3. Verify by recombining or by the coefficient checks (A + B = the x-coefficient). This calculator performs both modes — distinct and repeated linear factors — and prints the audit with the answer.

What is the Heaviside cover-up method?

Heaviside’s shortcut: to get the constant over (x − r), multiply the identity by (x − r) and substitute x = r — every other term vanishes, leaving that constant equal to (numerator at r)/(the remaining denominator factors at r). It works because the decomposition is an identity of polynomials, valid at every x including the roots where the original fractions are individually undefined. For repeated factors, cover-up yields the highest-power constant directly; the rest come from coefficient comparison (or differentiating the identity, Heaviside’s full method).

Why does a repeated factor need two fractions?

Because the decomposition must span all functions the denominator can produce: 1/(x − r)² alone cannot represent the (x − r)⁻¹ behavior that a general numerator creates. Concretely, solving (ax + b)/(x − r)² = B/(x − r)² only forces ax + b = B for all x — impossible unless a = 0. The staircase A/(x − r) + B/(x − r)² restores exactly the two degrees of freedom the numerator supplies. The rule generalizes: multiplicity m gets m terms, powers 1 through m.

What if the fraction is improper (numerator degree ≥ denominator)?

Divide first. (x³ + 1)/(x + 1) = x² − x + 1 + 0/(x+1) — polynomial part plus a proper remainder (here zero). The decomposition templates apply to the remainder only; the polynomial part carries through untouched (and integrates trivially). Our Polynomial Division Calculator performs that first step; this page assumes the proper form (ax + b over a quadratic) as its input contract states.

What about irreducible quadratic factors?

They take linear numerators: (2x + 3)/((x + 1)(x² + 4)) = A/(x + 1) + (Bx + C)/(x² + 4), solved by coefficient comparison (cover-up gives A instantly; B, C need the system). Their integrals produce a logarithm plus an arctangent — the reason the template differs. The widgets here cover the linear-factor cases (the most common coursework); the quadratic-factor template follows the identical principle: one fraction per factor, numerator one degree below the factor.

Where are partial fractions actually used?

Integrating rational functions (their primary calculus role — every rational integral reduces to logs and arctans through them), inverse Laplace transforms in engineering (transfer functions decompose into standard decay/oscillation terms), generating functions in combinatorics, and control theory’s pole analysis. The cover-up A-values are literally the residues of complex analysis — the same constants reappear as the residues in contour integration, a unification worth noticing when you meet it.

How do I check a decomposition?

Two fast audits, both printed by this calculator for the distinct case: A + B must equal the numerator’s x-coefficient a, and −A·r₂ − B·r₁ must equal its constant b. The universal audit: substitute any convenient x (try 0) into both the original fraction and your decomposition — equality at enough points proves the identity. Never integrate an unchecked decomposition; the checks cost twenty seconds and catch the sign errors that cost twenty points.

Is this partial fractions calculator free?

Yes — free, instant, no sign-up. Enter the numerator and the denominator’s roots for the distinct- or repeated-factor decomposition with cover-up steps and checks.