01 Key Concepts
What Partial Fractions Do
Rewrite a single complex fraction, like (5x+1)/((x+1)(x-2)), as a sum of simpler fractions with smaller denominators.
Setting Up the Decomposition
For distinct linear factors, assign one unknown constant to each factor's fraction: (5x+1)/((x+1)(x-2)) = A/(x+1) + B/(x-2).
Solving for the Unknown Constants
Multiply both sides by the original denominator to clear fractions, then substitute convenient x-values (or match coefficients) to solve for each constant.
Repeated Linear Factors
A repeated factor like (x-1)^2 requires a term for each power: A/(x-1) + B/(x-1)^2.
Why It's Useful
Simpler fractions are much easier to integrate, differentiate, or otherwise manipulate than one complicated combined fraction.
02 Key Formulas
- (px+q) / [(x-a)(x-b)] = A/(x-a) + B/(x-b)
03 Solved Examples
Example 1 Decompose (5x+1) / [(x+1)(x-2)] into partial fractions.
- Set up: 5x+1 = A(x-2) + B(x+1).
- Let x=-1 to eliminate B: 5(-1)+1 = A(-1-2) => -4 = -3A => A = 4/3.
- Let x=2 to eliminate A: 5(2)+1 = B(2+1) => 11 = 3B => B = 11/3.
Answer: (5x+1)/[(x+1)(x-2)] = (4/3)/(x+1) + (11/3)/(x-2)
Example 2 Decompose (3x+5) / [(x-1)(x+3)] into partial fractions.
- Set up: 3x+5 = A(x+3) + B(x-1).
- Let x=1: 3(1)+5 = A(1+3) => 8 = 4A => A=2.
- Let x=-3: 3(-3)+5 = B(-3-1) => -4 = -4B => B=1.
Answer: (3x+5)/[(x-1)(x+3)] = 2/(x-1) + 1/(x+3)
Example 3 Check the decomposition 2/(x-1) + 1/(x+3) by combining back into a single fraction and confirming the numerator.
- Common denominator: (x-1)(x+3).
- 2(x+3) + 1(x-1) = 2x+6+x-1 = 3x+5.
- This matches the original numerator, 3x+5.
Answer: Confirmed correct
04 Practice Questions
1Set up the partial fraction form for (2x+3)/[(x-1)(x+2)].
A/(x-1) + B/(x+2)
2For 5x/[(x-2)(x+1)] = A/(x-2)+B/(x+1), find A by letting x=2.
5(2)=A(2+1), 10=3A, A=10/3
3What kind of denominator does partial fraction decomposition break apart?
A product of simpler factors, usually linear
4Set up the partial fraction form for a repeated factor (x-4)^2.
A/(x-4) + B/(x-4)^2
5Why is partial fraction decomposition especially useful in calculus?
It makes integrating complicated rational expressions much easier
📄 Partial Fractions — Downloadable Worksheet
10 questions with a full answer key. Grab the PDF to print, or try the interactive version in your browser.