Improper Rational Functions
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
Partial fractions applies to proper rational functions, where the degree of the numerator is strictly less than the degree of the denominator. When the degree of the numerator is equal to or greater, the rational function is called improper, and partial fractions cannot be applied directly. The fix is straightforward: perform polynomial long division to extract a polynomial quotient, leaving a proper rational remainder. Then apply partial fractions to the remainder.
Quick Reference
When to divide: If $\deg(\text{numerator}) \geq \deg(\text{denominator})$, perform polynomial long division: $$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$$ where $S(x)$ is the quotient polynomial and $R(x)$ is the remainder with $\deg R < \deg Q$.
Integration: $\displaystyle\int\frac{P(x)}{Q(x)}\,dx = \int S(x)\,dx + \int\frac{R(x)}{Q(x)}\,dx$.
The first integral is a polynomial; the second is a proper rational function, handled by partial fractions.
Key Concepts
1. Recognizing an Improper Rational Function
Compare the degrees:
- $\dfrac{x^3 - 2}{x^2 - 1}$: degree 3 numerator, degree 2 denominator. Improper -- divide first.
- $\dfrac{x^2 + 1}{x^3 - x}$: degree 2 numerator, degree 3 denominator. Proper -- go straight to partial fractions.
2. Polynomial Long Division
Divide $P(x)$ by $Q(x)$ the same way you divide integers. Subtract multiples of $Q(x)$ from $P(x)$ until the remainder has degree less than $\deg Q$.
Example. Divide $x^3$ by $x^2 - 1$:
$x^3 \div (x^2 - 1)$: first term is $x\cdot(x^2 - 1) = x^3 - x$. Subtract: $x^3 - (x^3 - x) = x$. The remainder $x$ has degree 1, which is less than 2. So:
$$\frac{x^3}{x^2-1} = x + \frac{x}{x^2-1}.$$
3. Integrating After Division
The polynomial part $S(x)$ integrates directly by the power rule. The proper remainder $R(x)/Q(x)$ is integrated using partial fractions.
“The degree of the numerator must be exactly one more than the degree of the denominator for this to apply.” Long division applies whenever $\deg P \geq \deg Q$, regardless of the gap. If $\deg P = \deg Q$, the quotient is a nonzero constant. If $\deg P = \deg Q + 2$, the quotient is linear. The degree relationship determines the form of $S(x)$, not whether division is needed.
Worked Example
Evaluate $\displaystyle\int\frac{x^4-x^3+x^2-3x+2}{x^2-x+1}\,dx$.
Step 1 -- Divide. $\deg(\text{numerator}) = 4 \geq \deg(\text{denominator}) = 2$.
Divide $x^4 - x^3 + x^2 - 3x + 2$ by $x^2 - x + 1$:
- $x^4 \div x^2 = x^2$. Subtract $x^2(x^2 - x + 1) = x^4 - x^3 + x^2$. Remainder: $-3x + 2$.
- Now $\deg(-3x+2) = 1 < 2 = \deg(x^2-x+1)$. Stop.
$$\frac{x^4-x^3+x^2-3x+2}{x^2-x+1} = x^2 + \frac{-3x+2}{x^2-x+1}.$$
Step 2 -- Check the remainder. The denominator $x^2-x+1$ has discriminant $1-4 = -3 < 0$, so it is irreducible. The remainder $(-3x+2)/(x^2-x+1)$ is a proper fraction with an irreducible quadratic denominator.
Step 3 -- Integrate the polynomial part. $\displaystyle\int x^2\,dx = \dfrac{x^3}{3} + C_1$.
Step 4 -- Integrate the rational part. The derivative of $x^2-x+1$ is $2x-1$. Write $-3x+2 = -\frac{3}{2}(2x-1) + \frac{1}{2}$.
$\displaystyle\int\frac{-3x+2}{x^2-x+1}\,dx = -\frac{3}{2}\int\frac{2x-1}{x^2-x+1}\,dx + \frac{1}{2}\int\frac{1}{x^2-x+1}\,dx$.
First piece: $-\dfrac{3}{2}\ln(x^2-x+1)$.
Second piece: complete the square: $x^2-x+1 = (x-\frac{1}{2})^2 + \frac{3}{4}$.
$\dfrac{1}{2}\int\dfrac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}} = \dfrac{1}{2}\cdot\dfrac{1}{\sqrt{3}/2}\arctan\!\left(\dfrac{x-\frac{1}{2}}{\sqrt{3}/2}\right) = \dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{2x-1}{\sqrt{3}}\right)$.
Step 5 -- Combine.
$$\frac{x^3}{3} - \frac{3}{2}\ln(x^2-x+1) + \frac{1}{\sqrt{3}}\arctan\!\left(\frac{2x-1}{\sqrt{3}}\right) + C.$$
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Skipping division when numerator degree is equal to denominator degree | Directly decomposing $\frac{x^2}{x^2-1}$ | $\frac{x^2}{x^2-1} = 1 + \frac{1}{x^2-1}$; the leading term is $1$, not zero |
| Forgetting the remainder and integrating only the quotient | Integrating $x$ but forgetting $\frac{x}{x^2-1}$ | Both the quotient and the remainder must be integrated |
| Division arithmetic errors | Subtracting incorrectly during polynomial long division | Subtract the entire product (each coefficient), not just the leading term |
Leveled Practice
Level 1 -- Division Check
Problem 1. Without computing, state whether each rational function is proper or improper.
(a) $\dfrac{x^2+3}{x^3-1}$ (b) $\dfrac{x^3+x}{x^2-4}$ (c) $\dfrac{5}{(x-1)(x+2)}$
Show answer
(a) Degree 2 numerator, degree 3 denominator: proper. No division needed.
(b) Degree 3 numerator, degree 2 denominator: improper. Divide first.
(c) Degree 0 numerator, degree 2 denominator: proper. No division needed.
Level 2 -- Divide and Integrate
Problem 2. Evaluate $\displaystyle\int\frac{x^3+x}{x^2-1}\,dx$.
Show answer
Divide: $x^3 \div (x^2-1)$: quotient $x$, remainder $x^3 - x(x^2-1) = x$. Wait -- include the $+x$ in the numerator:
$x^3 + x = x(x^2-1) + 2x$, so $\dfrac{x^3+x}{x^2-1} = x + \dfrac{2x}{x^2-1}$.
$\dfrac{2x}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$. Cover-up: $A = 1$, $B = 1$.
$\displaystyle\int\left[x + \frac{1}{x-1} + \frac{1}{x+1}\right]dx = \frac{x^2}{2} + \ln|x-1| + \ln|x+1| + C = \frac{x^2}{2} + \ln|x^2-1| + C$.
Mastery Checklist
Mental Model
Partial fractions is like factoring out a common denominator in reverse. But you can only do this reverse step when the fraction is proper. An improper fraction is like a mixed number in arithmetic: $\frac{7}{3} = 2 + \frac{1}{3}$. Long division extracts the whole-number part (here: the polynomial $S(x)$) and leaves a proper fraction (here: the rational remainder $R/Q$). Only then can you apply the reverse-factoring step.
Connections
Looking back
- Polynomial long division: The arithmetic tool; review it if the mechanics are rusty.
- Partial fractions -- distinct and repeated linear (Section 7.4): Applied to the proper remainder.
Looking ahead
- Integration strategy (Section 7.5): The first step in integrating any rational function is checking whether long division is needed.
Back to Techniques of Integration | Next: Integration Strategy