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Math101learn.math101.caPartial Fractions
A rigorous, example-driven guide to partial fractions, including hypotheses, method choice, verification, and practice.
The central idea
A proper rational function can be decomposed into simpler fractions after its denominator is factored over the chosen field. Distinct linear factors use $A/(x-a)$; repeated factors require every power; irreducible quadratic factors use linear numerators. An improper rational function must first undergo polynomial division.
Definitions, hypotheses, and notation
Coefficient values can be solved by evaluating at roots only for the portions those roots isolate; repeated factors and irreducible quadratics often require derivative or coefficient comparison steps. Factoring over the reals keeps quadratics such as $x^2+1$ intact, while a complex decomposition would look different.
Domain intervals matter after integration. Logarithmic constants can differ across disconnected intervals separated by denominator zeros. Absolute values preserve the derivative $1/(x-a)$ on either side, but the original rational function still remains undefined at the pole.
Conceptual meaning
The decomposition rewrites one complicated rational function as a sum whose antiderivatives are logarithmic, reciprocal-power, or inverse-trigonometric forms. It is an algebraic identity on the common domain, not an approximation.
A dependable method and decision rule
- Compare degrees and perform long division if necessary.
- Factor the denominator completely.
- Write the full decomposition template, including repeated powers.
- Clear denominators and solve coefficients by strategic substitution or coefficient comparison.
- Integrate each term and differentiate to verify.
Fully worked example
Graphical or geometric meaning
Each denominator factor marks a singular location. The logarithmic antiderivative retains those same domain breaks. Coefficients describe the strength and sign of each simple pole's contribution.
Common mistakes and why they fail
Verification and reasonableness checks
- Recombine the fractions over the original denominator.
- Compare the resulting numerator coefficient by coefficient.
- Differentiate the integrated result on each domain interval.
Factor completely before choosing numerators
Make the rational function proper using polynomial division, then factor the denominator. A distinct linear factor gets a constant numerator; a repeated linear factor requires every power. An irreducible quadratic gets a linear numerator, and repeated quadratics likewise require all powers. Omitting a term makes the representation incomplete. After clearing denominators, solve coefficients using convenient roots and coefficient comparison, then recombine to verify. Linear denominators integrate to logarithms. Quadratics often require completing the square and separating a derivative-of-denominator contribution from an arctangent contribution. Domain exclusions from the original denominator remain even if later algebra cancels a factor, so state the interval on which an antiderivative formula applies.
Coefficient equations can be checked cheaply before integration. Substitute several ordinary values of $x$ into both rational expressions after avoiding denominator zeros; agreement at those points supports the algebra, while exact recombination supplies the definitive identity check.
Practice
- Decompose $1/[x(x+1)]$.
- What template corresponds to $(x-1)^2$?
- What comes before decomposition if numerator degree is larger?
Answers and brief solutions
- $1/x-1/(x+1)$.
- $A/(x-1)+B/(x-1)^2$.
- Polynomial division.
Connections and next steps
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Which is the correct decomposition of 1/[x(x+1)]?
- Set x=0 to get A=1.
- Set x=−1 to get B=−1.
- Thus 1/[x(x+1)]=1/x−1/(x+1).
End of lesson
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