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Partial fraction decomposition writes a proper rational function with distinct linear denominator factors as a sum of simple fractions.
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Formula
Algebraically dimensionless; x and all roots ri must use the same quantity and scale.
| Symbol | Meaning | Unit |
|---|---|---|
| Numerator polynomial with degree lower than Q | polynomial value | |
| Denominator polynomial with distinct real roots ri | polynomial value | |
| Root of the linear factor x−ri | same unit as x | |
| Constant coefficient of each partial fraction | depends on scaling and polynomial degrees |
\frac{P(x)}{Q(x)}=\sum_{i=1}^{m}\frac{A_i}{x-r_i}This compact form applies when a proper rational function has a denominator that factors into distinct linear terms. Repeated linear factors require terms such as A1/(x−r)+A2/(x−r)^2; irreducible quadratic factors take linear numerators. If deg(P)≥deg(Q), perform polynomial division first. Determine coefficients by equating coefficients or substituting convenient x values. Integrating the simple linear terms gives logarithms; the decomposition is valid only away from zeros of the original denominator.
This compact form applies when a proper rational function has a denominator that factors into distinct linear terms. Repeated linear factors require terms such as A1/(x−r)+A2/(x−r)^2; irreducible quadratic factors take linear numerators. If deg(P)≥deg(Q), perform polynomial division first. Determine coefficients by equating coefficients or substituting convenient x values. Integrating the simple linear terms gives logarithms; the decomposition is valid only away from zeros of the original denominator.
Partial fraction decomposition writes a proper rational function with distinct linear denominator factors as a sum of simple fractions.
Coefficients in the example
The decomposition applies for x≠−1,−2; repeated or quadratic factors need additional terms.
Rework the example for “Partial Fraction Decomposition of Rational Functions” and explain why the check is valid.
Decompose (5x+1)/((x+1)(x+2))=A/(x+1)+B/(x+2).
Equating coefficients gives A+B=5 and 2A+B=1, so A=−4 and B=9.
Check: −4(x+2)+9(x+1)=5x+1. Thus the original fraction is −4/(x+1)+9/(x+2), for x≠−1,−2.
Ignoring an assumption of the formula: This compact form applies when a proper rational function has a denominator that factors into distinct linear terms
Check units and model conditions before substitution; then verify the result in the original relation or a limiting case.
Formula cluster
Related relations in the formula sheet: grundintegrale-stammfunktion, integrationsregel-substitution, quotientenregel-ableitung.
Decompose (5x+1)/((x+1)(x+2))=A/(x+1)+B/(x+2). Equating coefficients gives A+B=5 and 2A+B=1, so A=−4 and B=9. Check: −4(x+2)+9(x+1)=5x+1. Thus the original fraction is −4/(x+1)+9/(x+2), for x≠−1,−2.
Reduce rational integrands to basic antiderivatives and compute inverse Laplace transforms with simple poles.
Here is how to work through a typical Partial Fraction Decomposition of Rational Functions (P(x)/Q(x) = Σ Ai/(x − ri)) task step by step:
Task
Rework the example for “Partial Fraction Decomposition of Rational Functions” and explain why the check is valid.
Solution path
Decompose (5x+1)/((x+1)(x+2))=A/(x+1)+B/(x+2).
Equating coefficients gives A+B=5 and 2A+B=1, so A=−4 and B=9.
Check: −4(x+2)+9(x+1)=5x+1. Thus the original fraction is −4/(x+1)+9/(x+2), for x≠−1,−2.
It decomposes a rational expression into simpler fractions whose denominators contain individual factors of the original denominator. For P(x)/Q(x), with deg(P)<deg(Q) and distinct linear factors x−ri, the form is ΣAi/(x−ri). This reveals the structure and can simplify integration or an inverse transform. The denominator must actually factor this way over the number system being used. Repeated factors and irreducible quadratic factors require corresponding additional terms. Determine the constants by coefficient matching and verify them by recombining the fractions. Check the algebraic identity.
Set (5x+1)/((x+1)(x+2)) equal to A/(x+1)+B/(x+2). Multiplying by both denominator factors gives 5x+1=A(x+2)+B(x+1). Match the x-coefficients and constant terms: A+B=5 and 2A+B=1. Subtraction yields A=−4, then B=9. Recombining gives −4(x+2)+9(x+1)=5x+1. This recovers the numerator and denominator of the original expression; x=−1 and x=−2 remain excluded. Alternatively, set x=−1: the B term vanishes, giving A=−4. At x=−2 the A term vanishes, so B=9. These substitutions follow after clearing denominators; the poles themselves remain excluded. This checks both solved coefficients. This makes the coefficient system transparent.
A repeated linear factor (x−r)ᵐ needs every power term: A1/(x−r)+A2/(x−r)2+…+Am/(x−r)ᵐ. For a quadratic factor ax2+bx+c that cannot be factored over the reals, use a linear numerator Cx+D. These forms account for possible remainders after polynomial division. If the numerator degree is at least the denominator degree, divide the polynomials first. Then determine coefficients by matching polynomial coefficients or substituting convenient values. Convenient roots can isolate individual coefficients when several factors occur. Choose a numerator of lower degree than each factor.
A simple term A/(x−r) has antiderivative A ln|x−r|. If a rational integrand is decomposed into such terms, integrate term by term and check the result by differentiation. Irreducible quadratic denominators can also produce logarithmic or arctangent terms, depending on the numerator. Decomposition prepares the expression for standard antiderivatives. For a definite integral, check whether the original denominator vanishes inside the interval: a pole can make the integral improper or cause it to diverge. Add the usual constant of integration afterward.
Check the degrees first: an improper rational function needs polynomial division before the partial fraction setup. Factor the denominator completely and record repeated factors; omitting a required power term makes the setup incomplete. Verify coefficients by recombining the fractions. Zeros of the original denominator remain excluded from the domain, even if a factor cancels. For the example, the identity −4(x+2)+9(x+1)=5x+1 checks the numerator. This substitution catches sign errors quickly. Also watch signs when expanding; recombination confirms that every term is correct.
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Curated exam set for "Partial Fraction Decomposition of Rational Functions":
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Which formula describes Partial Fraction Decomposition of Rational Functions?
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How do you rearrange Partial Fraction Decomposition of Rational Functions for Coefficients in the example?
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Which common mistake happens with Partial Fraction Decomposition of Rational Functions?
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