Adding fractions is a routine skill. You know that combines into . Partial fractions run that computation backwards, starting from the combined fraction and recovering the simple pieces. Calculus wants the pieces, because each one is easy to integrate and, in series, neighboring pieces often cancel.
Key terms
Proper rational expression
Definition A rational expression , with and polynomials and not the zero polynomial, is proper when is zero or the degree of is less than the degree of . Partial fraction decomposition applies only to proper expressions. If the expression is improper, first divide to get with zero or of lower degree than , and then decompose the proper part .
In plain words The top must have smaller degree than the bottom before you start. If it does not, long division comes first.
Example The expression is improper. Division gives , so
Only the proper part is split further. It equals , as the next sections show how to find.
Common denominator
Definition To add rational expressions, rewrite each over a common denominator, usually the product of the distinct factors to their highest powers, and add the numerators. For nonzero polynomials and ,
valid wherever and are both nonzero. Recombining over a common denominator is how every partial fraction result is verified.
In plain words Bring the pieces back to one fraction. If you get the original expression, the decomposition is right.
Example Over the common denominator ,
Linear factors
Definition A linear factor of the denominator is a factor with real . In a decomposition, a linear factor that appears once contributes one term with a constant numerator . A linear factor that appears to the power contributes terms, one for each power,
In plain words Every linear factor gets a constant on top. A repeated factor gets a separate term for each of its powers, not just the highest one.
Example For the repeated factor in use
Multiplying by gives . Matching the terms gives , and then gives . So the expression equals for .
Irreducible quadratic factors
Definition A quadratic with real coefficients and is irreducible over the reals when it has no real roots, that is when . Such a factor cannot be split into real linear factors. In a decomposition, an irreducible quadratic that appears once contributes one term with a numerator of degree at most one,
A repeated irreducible quadratic contributes one such term for each power, as with linear factors.
In plain words A quadratic that will not factor stays whole, and it gets a linear numerator, not just a constant, because a constant alone is not flexible enough.
Example The quadratic has discriminant , so it is irreducible. The expression therefore takes the form
By contrast, has discriminant and splits as , so it must be factored before choosing the form.
Unknown coefficients and coefficient matching
Definition The numerators , , and so on in the chosen form are unknown coefficients. Multiplying both sides by the original denominator gives an equation between two polynomials. Coefficient matching sets the coefficients of each power of equal on both sides, which gives a system of linear equations for the unknowns. Substituting convenient values of , such as the roots of linear factors, gives equations for the same unknowns and is often faster.
In plain words Clear the denominators, then compare the two sides power by power. Each power gives one equation.
Example For the form above, multiply by ,
The right side expands to . Matching powers gives three equations.
| Power | Equation |
|---|---|
| constant |
Adding the first and third and subtracting the second gives , so , then and . Setting is a quick cross check, since it gives , so again. Hence
Partial fraction decomposition
Definition Let be a proper rational expression with real coefficients, and factor into real linear factors and irreducible quadratic factors. Its partial fraction decomposition writes it as a sum of terms, one term with a real constant for each power up to the multiplicity of each linear factor and one term with real constants and for each power up to the multiplicity of each irreducible quadratic factor. Each term has its own unknown constants. Such a decomposition always exists and its coefficients are unique. Each term is called a partial fraction.
In plain words One fraction with a complicated denominator becomes a sum of fractions whose denominators are the simple pieces of the original.
Example Split into . Clearing denominators gives
At this reads , so . At it reads , so . Matching coefficients agrees, since and . The result is , which the common denominator section already confirmed.
In series the same idea gives the telescoping identity
Domain restriction
Definition A partial fraction decomposition is an equality of fractions that holds exactly where the original denominator is nonzero. Multiplying both sides by gives an equation between two polynomials that is true at every real except the finitely many roots of . The difference of the two sides is then a polynomial with infinitely many roots, and the only such polynomial is the zero polynomial. So the cleared equation is a polynomial identity, true for every real , roots of included. That is why substituting a root of the denominator into the cleared equation is legitimate, even though the fractions themselves are undefined there.
In plain words The fractions live only where the denominator is not zero. The cleared polynomial equation is true everywhere, so you may plug the forbidden values into that equation, but never into the fractions.
Example In the decomposition of the fractions are undefined at and . Substituting into is fine, because that equation between polynomials holds at every real . A check at an allowed value such as gives on the left and on the right.
Check yourself
- Decompose into partial fractions and state where the equality holds.
Show the answer
Factor and write . Setting gives and setting gives , so
- Write the form of the decomposition of without solving for the coefficients.
Show the answer
The quadratic has discriminant , so it is irreducible. The repeated linear factor needs a term for each power, so the form is
- Decompose and verify by recombining.
Show the answer
Write . Setting gives , so . Matching the terms gives , so , and matching constants gives , so . Thus
Recombining, , which is the original numerator.