Partial Fraction Decomposition Calculator
Decompose rational functions into partial fractions.
Supports distinct linear, repeated linear, and irreducible quadratic factors for integration and inverse Laplace transforms.
What Is Partial Fraction Decomposition? Partial fraction decomposition rewrites a rational function P(x)/Q(x) as a sum of simpler fractions. It is the algebraic reverse of combining fractions over a common denominator. The method is essential for: integrating rational functions, computing inverse Laplace transforms, and solving differential equations. It was systematically developed in the 18th century and appears in courses from pre-calculus through graduate-level engineering.
Requirements The degree of the numerator must be less than the degree of the denominator. If not, perform polynomial long division first to get a polynomial plus a proper fraction. The denominator Q(x) must be factored completely (over the reals).
Cases in Partial Fractions Case 1 — Distinct linear factors: Q(x) = (x−a₁)(x−a₂)…(x−aₙ) P(x)/Q(x) = A₁/(x−a₁) + A₂/(x−a₂) + … + Aₙ/(x−aₙ)
Case 2 — Repeated linear factors: factor (x−a)^n Contributes: A₁/(x−a) + A₂/(x−a)² + … + Aₙ/(x−a)^n
Case 3 — Irreducible quadratic factors: (x²+bx+c) where b²−4c < 0 Contributes: (Ax+B)/(x²+bx+c)
Case 4 — Repeated irreducible quadratics: (x²+bx+c)^n Each power contributes a term (Aₖx+Bₖ)/(x²+bx+c)^k for k=1..n
Finding the Coefficients Method 1 — Substitution: multiply both sides by Q(x), then substitute x = root of each factor. Method 2 — Expanding and matching: expand the right side and equate coefficients of each power of x. Method 3 — Heaviside cover-up (for distinct linear factors): cover the factor (x−aᵢ) in the denominator and evaluate at x = aᵢ.
Integration Using Partial Fractions ∫ A/(x−a) dx = A·ln|x−a| + C ∫ A/(x−a)^n dx = A·(x−a)^(1−n)/(1−n) + C (n ≠ 1) ∫ (Ax+B)/(x²+bx+c) dx → complete the square, then use arctan form
Example: (3x+1)/[(x−1)(x+2)] Set up: A/(x−1) + B/(x+2) = (3x+1)/[(x−1)(x+2)] Multiply: A(x+2) + B(x−1) = 3x+1 Set x=1: 3A = 4 → A = 4/3 Set x=−2: −3B = −5 → B = 5/3 Result: (4/3)/(x−1) + (5/3)/(x+2)
Laplace Transform Applications Inverse Laplace transforms require partial fractions to convert algebraic expressions back to time-domain functions. L⁻¹{1/(s−a)} = eᵃᵗ, L⁻¹{1/(s²+ω²)} = sin(ωt)/ω, L⁻¹{s/(s²+ω²)} = cos(ωt). Control system transfer functions in s-domain are routinely inverted via partial fractions.