Partial Fractions: Repeated Linear Factors
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.4: “Partial Fractions” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-4-partial-fractions |
| Textbook used in class | Stewart, Calculus, Section 7.4: “Integration of Rational Functions by Partial Fractions” |
Opening Scenario
When the denominator has a repeated factor $(x - r)^n$, one term $\dfrac{A}{x-r}$ is not enough to represent the decomposition. The rational function $\dfrac{1}{(x-1)^2}$ cannot be written as $\dfrac{A}{x-1}$ for any constant $A$. A second term $\dfrac{B}{(x-1)^2}$ is needed.
The pattern generalizes: a factor $(x-r)^n$ in the denominator contributes $n$ terms to the partial fraction decomposition, one for each power from 1 to $n$.
Quick Reference
Partial fractions for a repeated linear factor $(ax+b)^n$:
$$\frac{P(x)}{(ax+b)^n\cdots} = \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_n}{(ax+b)^n} + \cdots$$
Integration: Each term $\dfrac{A_k}{(ax+b)^k}$ integrates as follows:
- $k = 1$: $\dfrac{A_1}{a}\ln|ax+b| + C$.
- $k \geq 2$: $\dfrac{A_k}{a}\cdot\dfrac{(ax+b)^{1-k}}{1-k} + C$ (power rule in $ax+b$).
Key Concepts
1. Why More Terms Are Needed
If $(x-r)^n$ appears in the denominator, the general “contribution” from this factor to the partial fraction decomposition is: $$\frac{A_1}{(x-r)} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_n}{(x-r)^n}.$$
This can be verified: adding these terms over the common denominator $(x-r)^n$ yields a numerator of degree up to $n-1$, which can represent any polynomial of that degree in $A_1, \ldots, A_n$. One term alone has a constant numerator (degree 0), which is insufficient when the true numerator restricted to this factor has higher degree.
2. Finding the Constants
The cover-up method still finds $A_n$ (the highest-power constant): multiply both sides by $(x-r)^n$ and set $x = r$. The other constants $A_1, \ldots, A_{n-1}$ are found by substituting other values of $x$ or by equating polynomial coefficients.
Example setup. $\dfrac{x^2+3}{(x-1)^2(x+2)}$:
$$\frac{x^2+3}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.$$
$B$: cover-up at $x = 1$: $\dfrac{1+3}{3} = \dfrac{4}{3}$, so $B = \dfrac{4}{3}$.
$C$: cover-up at $x = -2$: $\dfrac{4+3}{(-3)^2} = \dfrac{7}{9}$, so $C = \dfrac{7}{9}$.
$A$: substitute a convenient value, e.g., $x = 0$: $\dfrac{3}{(-1)^2(2)} = \dfrac{A}{-1} + \dfrac{B}{1} + \dfrac{C}{2}$, giving $\dfrac{3}{2} = -A + \dfrac{4}{3} + \dfrac{7}{18}$. Solve for $A$.
“A repeated factor $(x-r)^2$ just requires one term $\frac{A}{(x-r)^2}$.” A factor $(x-r)^2$ requires two terms: $\dfrac{A}{x-r} + \dfrac{B}{(x-r)^2}$. Writing only the highest power misses the $A/(x-r)$ term. The error becomes apparent when you try to clear denominators and find the equations have no solution.
Worked Example
Evaluate $\displaystyle\int\frac{x}{(x-1)^2(x+1)}\,dx$.
Step 1 -- Decompose.
$$\frac{x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}.$$
Step 2 -- Clear denominators. $x = A(x-1)(x+1) + B(x+1) + C(x-1)^2$.
$x = 1$: $1 = 2B \Rightarrow B = \frac{1}{2}$.
$x = -1$: $-1 = 4C \Rightarrow C = -\frac{1}{4}$.
$x = 0$: $0 = A(-1)(1) + B(1) + C(1) = -A + \frac{1}{2} - \frac{1}{4} \Rightarrow A = \frac{1}{4}$.
Step 3 -- Integrate.
$$\int\left[\frac{1/4}{x-1} + \frac{1/2}{(x-1)^2} + \frac{-1/4}{x+1}\right]dx.$$
$= \frac{1}{4}\ln|x-1| + \frac{1}{2}\cdot\frac{(x-1)^{-1}}{-1} + \left(-\frac{1}{4}\right)\ln|x+1| + C$
$= \frac{1}{4}\ln|x-1| - \frac{1}{2(x-1)} - \frac{1}{4}\ln|x+1| + C$.
Boxed answer: $\dfrac{1}{4}\ln\!\left|\dfrac{x-1}{x+1}\right| - \dfrac{1}{2(x-1)} + C$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| One term for a repeated factor | $\frac{A}{(x-1)^2}$ for the factor $(x-1)^2$ | Use two terms: $\frac{A}{x-1} + \frac{B}{(x-1)^2}$ |
| Using cover-up on intermediate powers | Trying cover-up ($x=1$) to find $A$ in $\frac{A}{x-1}$ | Cover-up at $x=1$ gives only the highest-power constant $B$; use substitution or coefficient matching for $A$ |
| Integrating $\frac{1}{(x-r)^2}$ as a logarithm | Writing $\int\frac{1}{(x-r)^2}\,dx = \ln|(x-r)^2| + C$ | The correct antiderivative is $\frac{-1}{x-r} + C$ (power rule) |
Leveled Practice
Level 1 -- Setup
Problem 1. Write the correct partial fraction form for $\dfrac{3x+1}{x^2(x-2)}$.
Show answer
$x^2 = x\cdot x$ is a repeated linear factor with $r = 0$, multiplicity 2.
$$\frac{3x+1}{x^2(x-2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2}.$$
Level 2 -- Full Computation
Problem 2. Evaluate $\displaystyle\int\frac{1}{(x-1)(x-2)^2}\,dx$.
Show answer
$\dfrac{1}{(x-1)(x-2)^2} = \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{(x-2)^2}$.
Clear: $1 = A(x-2)^2 + B(x-1)(x-2) + C(x-1)$.
$x=1$: $1 = A \Rightarrow A = 1$.
$x=2$: $1 = C \Rightarrow C = 1$.
$x=0$: $1 = 4A + 2B - C = 4 + 2B - 1 \Rightarrow B = -1$.
$\displaystyle\int\left[\frac{1}{x-1} - \frac{1}{x-2} + \frac{1}{(x-2)^2}\right]dx = \ln|x-1| - \ln|x-2| - \frac{1}{x-2} + C$.
Mastery Checklist
Mental Model
A repeated factor $(x-r)^n$ in the denominator represents a root of multiplicity $n$. The partial fraction decomposition needs to account for all $n$ “layers” of that root: terms $\frac{A_1}{x-r}$ through $\frac{A_n}{(x-r)^n}$. Omitting any layer leaves the system underdetermined, and the missing layer’s contribution ends up lumped into the others, making the constants wrong.
Connections
Looking back
- Distinct linear factors (Section 7.4): The building block; this lesson extends it to repeated factors.
Looking ahead
- Irreducible quadratic factors (Section 7.4): The final case, for factors with no real roots.
Back to Techniques of Integration | Next: Irreducible Quadratic Factors