Why Repeated Factors Need Special Attention
A rational function can have a denominator containing the same linear factor more than once. The repeated factor affects the form of its partial-fraction decomposition: a factor with multiplicity three requires terms for the first, second, and third powers. Those terms do not all integrate in the same way. The first-power term produces a logarithm, while higher powers produce negative powers.
The essential task is to include the full sequence of required powers before finding coefficients. Leaving one out can make a correct decomposition impossible, even if the remaining terms look plausible.
Setting Up the Decomposition
For a proper rational function, the degree of the numerator is less than the degree of the denominator. If a rational function is not proper, polynomial division must be done first. When the denominator consists of linear factors, each distinct factor contributes its own terms. A factor \((x-r)^m\) contributes all powers from \((x-r)^{-1}\) through \((x-r)^{-m}\).
For example, a proper rational function with denominator \((x-1)^2(x+2)\) has the form
To determine the constants, multiply both sides by the original denominator. The result is a polynomial identity, so the coefficients of matching powers of \(x\) must be equal. The earlier material on partial fractions with distinct linear factors uses the same clearing-denominators idea; the additional feature here is the complete set of powers for each repeated factor.
Why Every Power Is Included
The need for the full sequence can be understood from the algebra after clearing denominators. Multiplying by \((x-r)^m\) turns the contribution from the repeated factor into a polynomial of the form
Each coefficient \(A_j\) controls a different power of \((x-r)\) in this polynomial. Omitting one of the terms forces the corresponding coefficient to be zero, which may not be consistent with the numerator of the original rational function. For instance, when \(m=3\), the full contribution is \(A_1(x-r)^2+A_2(x-r)+A_3\). The linear, constant, and quadratic parts all have a place in the identity. A particular coefficient may turn out to be zero, but the term should still be included when setting up the general form.
Integrating the Terms
Once the decomposition is found, integrate each term. For \(x\ne r\), the first power gives a logarithm. For powers greater than one, apply the power rule to \((x-r)^{-j}\). The antiderivative formulas are
The logarithm uses an absolute value because the real logarithm is defined for positive inputs, while \(x-r\) can be positive or negative. Each formula applies on an interval that does not contain \(r\); the rational function itself is undefined at \(x=r\).
To justify the second formula, write the integrand as \((x-r)^{-j}\). The derivative of \((x-r)^{1-j}\), by the chain rule, is \((1-j)(x-r)^{-j}\). Therefore multiplying \((x-r)^{1-j}\) by \(1/(1-j)\) gives an antiderivative. Since \(1/(1-j)=-1/(j-1)\), this is precisely the displayed formula. For the first formula, the derivative of \(\ln|x-r|\) is \(1/(x-r)\) wherever \(x\ne r\).
Worked Examples
Worked Example: A Repeated Factor of Order Two
Decompose and integrate \( \frac{3x+5}{(x-1)^2} \).
The denominator has a repeated factor of multiplicity two, so include both required terms:
Multiply both sides by \((x-1)^2\):
Match coefficients. The coefficient of \(x\) gives \(A=3\). The constant terms give \(5=-A+B\), so \(5=-3+B\), and \(B=8\). Thus
Integrate each term:
Check by differentiation: \( \frac{d}{dx}\left(3\ln|x-1|-8(x-1)^{-1}\right)=3/(x-1)+8/(x-1)^2 \). Combining these terms gives \( [3(x-1)+8]/(x-1)^2=(3x+5)/(x-1)^2 \), as required.
Worked Example: A Repeated Factor Alongside a Distinct Factor
Decompose and integrate \( \frac{3x^2-x-5}{(x-1)^2(x+2)} \).
The repeated factor \((x-1)^2\) requires two terms, and the distinct factor \(x+2\) requires one:
Clear the denominators:
Expand each product: \((x-1)(x+2)=x^2+x-2\), and \((x-1)^2=x^2-2x+1\). The right side becomes
Matching coefficients gives \(A+C=3\), \(A+B-2C=-1\), and \(-2A+2B+C=-5\). The values \(A=2\), \(B=-1\), \(C=1\) satisfy all three equations: \(2+1=3\), \(2-1-2=-1\), and \(-4-2+1=-5\). Therefore
Integrating gives
For the check, differentiate the proposed antiderivative to obtain \(2/(x-1)-1/(x-1)^2+1/(x+2)\). With the common denominator \((x-1)^2(x+2)\), the numerator is \(2(x-1)(x+2)-(x+2)+(x-1)^2\). Expanding gives \(2x^2+2x-4-x-2+x^2-2x+1=3x^2-x-5\). This recovers the original integrand.
Worked Example: A Repeated Factor of Order Three
Decompose and integrate \( \frac{x^2+4x}{(x+1)^3} \).
Since the repeated factor has multiplicity three, include all three powers:
After multiplying by \((x+1)^3\), the identity is
Expand the right side: \(A(x^2+2x+1)+Bx+B+C=Ax^2+(2A+B)x+(A+B+C)\). Matching coefficients gives \(A=1\), \(2A+B=4\), and \(A+B+C=0\). Thus \(B=2\), and \(1+2+C=0\) gives \(C=-3\). The decomposition is
Integrate term by term:
Differentiate to check: the derivative is \(1/(x+1)+2/(x+1)^2-3/(x+1)^3\). Putting these terms over \((x+1)^3\) gives numerator \((x+1)^2+2(x+1)-3=x^2+4x\). The original integrand is recovered.
A Common Mistake: Skipping an Intermediate Power
A frequent error is to include only the first and last powers for a repeated factor. For example, one might try to write
Clearing denominators by multiplying through by \((x-1)^3\) would require
The coefficient of \(x^2\) on the right is \(A\), so matching the left side forces \(A=0\). But then the right side is the constant \(C\), which cannot equal \(x-1\) for every \(x\). The proposed form is impossible.
The correction is to use the powers required by the denominator structure, including the missing second power. In this case, \(1/(x-1)^2\) is already the second-power term; when setting up a general decomposition for a denominator containing \((x-1)^3\), include terms with powers one, two, and three. Do not decide in advance that a term can be omitted; let coefficient matching show whether its coefficient is zero.
A Reliable Procedure
If its numerator degree is at least its denominator degree, perform polynomial division first.
For every repeated linear factor of multiplicity \(m\), include one term for each power from one through \(m\).
Clear denominators, expand, and match coefficients in the resulting polynomial identity.
Use a logarithm for a first power and the power rule for each higher power.
Differentiate the antiderivative and verify that it equals the original rational function.
Why the Structure Matters
The decomposition separates a complicated rational function into terms whose antiderivatives are familiar. The repeated factor determines which powers appear, and therefore whether each term produces a logarithm or a negative power. If even one required term is absent, coefficient matching may fail; if the decomposition is correct, integrating term by term and differentiating to check provides a direct way to confirm the result.
Check Your Understanding
Use the repeated-factor rules to answer these questions.
- What terms must be included for a repeated factor \((x-3)^4\)?
- Why does a term with denominator \((x-r)\) integrate differently from one with denominator \((x-r)^3\)?
- Set up, but do not solve, the partial-fraction form for a proper rational function with denominator \((x+1)^2(x-4)^3\).
- Integrate \(5/(x+2)^2\), and check your result by differentiation.
- Why should a coefficient be allowed to equal zero rather than omitting its term before coefficient matching?
