Partial Fractions: Distinct 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
Adding fractions $\dfrac{1}{x-2} + \dfrac{1}{x+3} = \dfrac{2x+1}{(x-2)(x+3)}$ is straightforward: find a common denominator and combine. Partial fractions is the reverse: given $\dfrac{2x+1}{(x-2)(x+3)}$, split it back into $\dfrac{1}{x-2} + \dfrac{1}{x+3}$.
Why does this help with integration? Because $\int\dfrac{1}{x-a}\,dx = \ln|x-a| + C$ is immediate, while $\int\dfrac{2x+1}{(x-2)(x+3)}\,dx$ is not obvious.
Quick Reference
Setup for distinct linear factors. If the denominator factors completely into distinct linear factors $(a_1 x + b_1)(a_2 x + b_2)\cdots(a_n x + b_n)$, write: $$\frac{P(x)}{(a_1 x+b_1)(a_2 x+b_2)\cdots(a_n x+b_n)} = \frac{A_1}{a_1 x+b_1} + \frac{A_2}{a_2 x+b_2} + \cdots + \frac{A_n}{a_n x+b_n}.$$
Integration: $\displaystyle\int\frac{A}{ax+b}\,dx = \frac{A}{a}\ln|ax+b| + C$.
Key Concepts
1. Precondition: Proper Fraction
Partial fractions applies to proper rational functions, where the degree of the numerator is strictly less than the degree of the denominator. If the degree of the numerator is equal to or greater than the degree of the denominator, perform polynomial long division first to obtain a proper fraction remainder.
2. The Method of Undetermined Coefficients
Multiply both sides of the partial fraction equation by the denominator to obtain a polynomial identity. Then find the constants $A, B, \ldots$ by:
- Cover-up method (Heaviside): Substitute the root $x = r$ of each linear factor on both sides (the other terms vanish because they share the same root in the denominator, after cancellation).
- Expanding and equating coefficients: Expand the right side, collect by powers of $x$, and match coefficients.
3. After Decomposition
Each term $\dfrac{A}{ax+b}$ integrates via: $$\int\frac{A}{ax+b}\,dx = \frac{A}{a}\ln|ax+b| + C.$$
“Each constant $A_i$ can be any real number, so there is no check.” The constants are uniquely determined by the identity (two polynomials agree at all $x$ if and only if their coefficients match). After finding $A, B, \ldots$, verify by recombining the partial fractions and checking they equal the original rational function.
Worked Example
Evaluate $\displaystyle\int\frac{3x+5}{(x-1)(x+2)}\,dx$.
Step 1 -- Partial fraction setup.
$$\frac{3x+5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$$
Step 2 -- Clear denominators.
$3x + 5 = A(x+2) + B(x-1)$.
Step 3 -- Find constants (cover-up).
$x = 1$: $3(1)+5 = A(3) \Rightarrow A = \dfrac{8}{3}$.
$x = -2$: $3(-2)+5 = B(-3) \Rightarrow -1 = -3B \Rightarrow B = \dfrac{1}{3}$.
Step 4 -- Integrate.
$$\int\left[\frac{8/3}{x-1} + \frac{1/3}{x+2}\right]dx = \frac{8}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C.$$
Boxed answer: $\dfrac{8}{3}\ln|x-1| + \dfrac{1}{3}\ln|x+2| + C$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Trying partial fractions on an improper fraction | Decomposing $\frac{x^3+1}{(x-1)(x+2)}$ directly | Degree 3 $\geq$ degree 2: perform long division first to get a polynomial plus a proper fraction |
| Forgetting the $\frac{1}{a}$ factor when integrating $\frac{A}{ax+b}$ | Writing $\int\frac{3}{2x+1}\,dx = 3\ln|2x+1| + C$ | $\int\frac{3}{2x+1}\,dx = \frac{3}{2}\ln|2x+1| + C$ |
| Assuming a factor $(x-r)^2$ is treated the same as $(x-r)$ | Writing $\frac{A}{x-r} + \frac{B}{x-r}$ | Repeated factors require $\frac{A}{x-r} + \frac{B}{(x-r)^2}$ (covered in the next lesson) |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Evaluate $\displaystyle\int\frac{1}{x(x-3)}\,dx$.
Show answer
$\dfrac{1}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}$.
Clear: $1 = A(x-3) + Bx$. $x=0$: $1 = -3A \Rightarrow A = -1/3$. $x=3$: $1 = 3B \Rightarrow B = 1/3$.
$\displaystyle\int\left[-\frac{1/3}{x} + \frac{1/3}{x-3}\right]dx = -\frac{1}{3}\ln|x| + \frac{1}{3}\ln|x-3| + C = \frac{1}{3}\ln\!\left|\frac{x-3}{x}\right| + C$.
Level 2 -- Three Factors
Problem 2. Evaluate $\displaystyle\int\frac{2}{(x-1)(x)(x+1)}\,dx$.
Show answer
$\frac{2}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.
$2 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$.
$x=0$: $2 = A(-1)(1) \Rightarrow A = -2$. $x=1$: $2 = B(1)(2) \Rightarrow B = 1$. $x=-1$: $2 = C(-1)(-2) \Rightarrow C = 1$.
$\displaystyle\int\left[-\frac{2}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right]dx = -2\ln|x| + \ln|x-1| + \ln|x+1| + C$.
Mastery Checklist
Mental Model
Adding fractions requires finding a common denominator; partial fractions is the reverse -- finding the “summands” from the combined fraction. The key insight is that any proper rational function with a completely factored denominator can be written as a sum of simple fractions, one per factor. Each simple fraction integrates to a logarithm, reducing the entire problem to a sum of basic integrals.
Connections
Looking back
- Logarithm antiderivative: $\int\frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C$ is the key integration fact.
Looking ahead
- Repeated linear factors (Section 7.4): A factor $(x-r)^n$ requires $n$ separate terms.
- Irreducible quadratic factors (Section 7.4): Factors $ax^2+bx+c$ (no real roots) require a different form.
Back to Techniques of Integration | Next: Repeated Linear Factors