Partial Fraction Decomposition Calculator
Decomposes a rational function with a factorable denominator into a sum of simpler partial fractions. Essential for evaluating integrals and performing inverse Laplace transforms.
About this calculator
Partial fraction decomposition rewrites a rational function P(x)/Q(x) — where the degree of P is less than the degree of Q — as a sum of simpler fractions. For a denominator with two distinct linear factors (x − a)(x − b), the decomposition is: P(x)/[(x−a)(x−b)] = A/(x−a) + B/(x−b). To find A and B, multiply both sides by (x−a)(x−b) and equate coefficients or substitute the roots x = a and x = b (the cover-up method). For example, if P(x) = mx + n, substituting x = a gives A = (ma + n)/(a − b) and x = b gives B = (mb + n)/(b − a). The formula used here computes one numerator coefficient via: A = (num_coeff_a · denom_factor2 + num_constant) / (denom_factor2 − denom_factor1). Repeated or irreducible quadratic factors require additional terms.
How to use
Decompose (5x + 3) / [(x − 1)(x − 4)]. Enter num_coeff_a = 5, num_constant = 3, denom_factor1 = 1, denom_factor2 = 4. Step 1 — compute A using x = 1: A = (5·1 + 3)/(1 − 4) = 8/(−3) = −8/3. Step 2 — compute B using x = 4: B = (5·4 + 3)/(4 − 1) = 23/3. Result: (5x + 3)/[(x−1)(x−4)] = (−8/3)/(x−1) + (23/3)/(x−4). Verify by recombining over the common denominator.
Frequently asked questions
How do you perform partial fraction decomposition with repeated linear factors?
When the denominator contains a repeated factor (x − a)², you must include two separate terms: A/(x − a) + B/(x − a)². Multiply through by (x − a)² and either substitute x = a to find B directly, then equate the x-coefficient to find A. For a factor (x − a)ⁿ, you need n terms with increasing powers in the denominator. This calculator handles the two-distinct-linear-factor case; repeated factors require the extended setup described above.
Why is partial fraction decomposition useful for integration in calculus?
Integrating a complex rational function directly is often intractable, but each partial fraction term has a standard integral: ∫ A/(x − a) dx = A·ln|x − a| + C. Decomposing first turns one hard integral into a sum of elementary logarithmic (or arctangent, for irreducible quadratics) integrals. This technique is indispensable in calculus II and in engineering for inverse Laplace transforms, where complex transfer functions must be broken into recognisable standard forms before table lookup.
What conditions must a rational function meet for partial fraction decomposition to work?
The degree of the numerator must be strictly less than the degree of the denominator (proper fraction). If it is not, perform polynomial long division first to extract the integer polynomial part, then decompose the proper remainder. The denominator must also be fully factored — into linear and/or irreducible quadratic factors over the reals. Each factor's type determines the form of the corresponding partial fraction term. Meeting these conditions guarantees a unique decomposition exists.