Partial fractions is a method for integrating rational functions $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. The idea is to break a complicated fraction into a sum of simpler fractions that are easy to integrate (like $\frac{A}{x-a}$ or $\frac{Ax+B}{x^2+bx+c}$). This is a BC Calculus topic.
🎯 In this section you will learn
- When to use partial fractions (denominator factors into linear or irreducible quadratic factors)
- How to set up partial fraction decomposition for different cases
- How to solve for the unknown coefficients
- How to integrate the resulting simpler fractions
📌 When to Use Partial Fractions
Partial fractions works when the denominator $Q(x)$ factors into linear factors or irreducible quadratic factors. The numerator $P(x)$ must have degree less than $Q(x)$ (if not, use polynomial long division first).
📋 The Four Cases
| Case | Denominator Factors | Partial Fraction Form |
|---|
| Case 1 | Distinct linear factors | $\frac{A}{x-a} + \frac{B}{x-b} + \cdots$ |
| Case 2 | Repeated linear factors | $\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k}$ |
| Case 3 | Distinct irreducible quadratics | $\frac{Ax+B}{x^2+bx+c} + \frac{Cx+D}{x^2+ex+f} + \cdots$ |
| Case 4 | Repeated irreducible quadratics | $\frac{A_1x+B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2} + \cdots$ |
💡 Important
Always perform polynomial long division first if the degree of the numerator is greater than or equal to the degree of the denominator.
Find $\displaystyle \int \frac{5x-1}{x^2 - x - 2} dx$.
①
Factor denominator
$x^2 - x - 2 = (x-2)(x+1)$
②
Set up partial fractions
$\frac{5x-1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$
③
Multiply through by denominator
$5x-1 = A(x+1) + B(x-2)$
④
Solve for $A$ and $B$
Let $x=2$: $5(2)-1 = A(3) + B(0)$ → $9 = 3A$ → $A=3$
Let $x=-1$: $5(-1)-1 = A(0) + B(-3)$ → $-6 = -3B$ → $B=2$
⑤
Rewrite and integrate
$\int \frac{5x-1}{(x-2)(x+1)} dx = \int \left(\frac{3}{x-2} + \frac{2}{x+1}\right) dx = 3\ln|x-2| + 2\ln|x+1| + C$
Find $\displaystyle \int \frac{x+2}{(x-1)^2} dx$.
①
Set up partial fractions
$\frac{x+2}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$
②
Multiply through
$x+2 = A(x-1) + B$
③
Solve for $A$ and $B$
Let $x=1$: $3 = B$
Expand: $x+2 = Ax - A + 3$ → Compare coefficients: $A=1$, $-A+3=2$ checks.
④
Integrate
$\int \frac{x+2}{(x-1)^2} dx = \int \frac{1}{x-1} dx + \int \frac{3}{(x-1)^2} dx = \ln|x-1| - \frac{3}{x-1} + C$
Find $\displaystyle \int \frac{2x+1}{x(x^2+1)} dx$.
①
Set up partial fractions
$\frac{2x+1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$
②
Multiply through
$2x+1 = A(x^2+1) + (Bx+C)x = A(x^2+1) + Bx^2 + Cx$
③
Expand and compare coefficients
$2x+1 = (A+B)x^2 + Cx + A$
Compare: $x^2$: $A+B=0$; $x$: $C=2$; constant: $A=1$
Then $1+B=0$ → $B=-1$
④
Integrate
$\int \frac{2x+1}{x(x^2+1)} dx = \int \frac{1}{x} dx + \int \frac{-x+2}{x^2+1} dx = \ln|x| - \frac{1}{2}\ln|x^2+1| + 2\arctan x + C$
Find $\displaystyle \int \frac{x^3}{x^2 - 1} dx$.
①
Long division
$\frac{x^3}{x^2-1} = x + \frac{x}{x^2-1}$
②
Partial fractions on remainder
$\frac{x}{x^2-1} = \frac{x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
$x = A(x+1) + B(x-1)$
Let $x=1$: $1 = 2A$ → $A=\frac{1}{2}$
Let $x=-1$: $-1 = -2B$ → $B=\frac{1}{2}$
③
Integrate
$\int \frac{x^3}{x^2-1} dx = \int x dx + \frac{1}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{1}{x+1} dx = \frac{x^2}{2} + \frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C$
⚠️ Common Mistakes
- Forgetting long division: If the degree of the numerator is ≥ degree of denominator, you must divide first.
- Incorrect partial fraction setup: Repeated factors require a term for each power. Irreducible quadratics require $Ax+B$ in the numerator.
- Missing absolute values: $\int \frac{1}{x} dx = \ln|x| + C$, not $\ln x + C$.
🔍 Key Takeaways
- Partial fractions decomposes $\frac{P(x)}{Q(x)}$ into simpler fractions.
- Four cases: distinct linear, repeated linear, distinct irreducible quadratics, repeated irreducible quadratics.
- Use polynomial long division first if the numerator's degree is ≥ denominator's degree.
- Solve for coefficients using substitution or comparing coefficients.
- Integrate using $\ln|\cdot|$, arctan, or power rule.
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