Frações Parciais 1/(ax+b)(cx+d)

Partial fractions of 1/((ax+b)(cx+d))

When the denominator is a product of two distinct linear factors, the rational function splits into 1/((ax + b)(cx + d)) = A/(ax + b) + B/(cx + d). The quickest way to pin down A and B is the Heaviside cover-up. Multiply through by (ax + b) and plug in x = −b/a, which leaves A = 1/(c·(−b/a) + d). The same trick with x = −d/c gives B = 1/(a·(−d/c) + b). A concrete case: 1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2). This only works while the factors stay distinct, that is a·d − b·c ≠ 0. Should they coincide, the denominator turns into (ax + b)² and you switch to a repeated-root form. The same idea extends to repeated factors, with one term per power, and to irreducible quadratics, where the numerator is Cx + D.

Where partial fractions are used

• Calculus — once you decompose it, ∫1/((ax + b)(cx + d)) dx becomes a sum of logarithms.
• Inverse Laplace transform — you lean on it to bring transfer functions back to time-domain responses in control engineering and circuits.
• Linear ODEs — Laplace-based solutions depend on breaking apart the rational function you end up with.
• Filter theory — once a transfer function is written as elementary terms, stability and frequency analysis get easier.

FAQ

What if a·d = b·c? Then the two factors are proportional, so the denominator collapses to (ax + b)² up to a constant. Reach for the repeated-root form A/(ax + b) + B/(ax + b)².

Why is the integral always a logarithm? Each A/(ax + b) integrates to (A/a) · ln|ax + b|, and adding those up leaves you with a sum of lns.

Does the cover-up always work? With simple, non-repeated linear factors it is exact. For repeated or quadratic factors, you'll want to combine it with comparison of coefficients.

What signs should I expect? Take a = c = 1 with b ≠ d. You land on A = 1/(d − b) and B = 1/(b − d) = −A, so the two coefficients share a magnitude but flip sign.