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Partial Fractions

Partial Fractions

The process of writing any proper rational expression as a sum of proper rational expressions. This method is use in integration as shown below.

Note: Improper rational expressions can also be rewritten using partial fractions. You must, however, use polynomial long division first before finding a partial fractions representation.

 

Example 1: Partial fractions decomposition of (4x+1)/(x²-x-2) = 1/(x+1) + 3/(x-2), solved by substituting x=2 and x=-1.


Example 2: Partial fractions integration of (4x+1)/(x²−x−2) = ln|(x+1)(x−2)³|+C

 

 

See also

Integration methods

Key Formula

P(x)(x−a)(x−b)=Ax−a+Bx−b\frac{P(x)}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}
Where:
  • P(x)P(x) = A polynomial in the numerator with degree less than the degree of the denominator
  • a,ba, b = Distinct real roots of the denominator
  • A,BA, B = Unknown constants to be determined

Worked Example

Problem: Decompose the rational expression into partial fractions: 5x+3(x+1)(x−2)\frac{5x + 3}{(x + 1)(x - 2)}
Step 1: Set up the partial fraction form. Since the denominator has two distinct linear factors, write one fraction for each factor.
5x+3(x+1)(x−2)=Ax+1+Bx−2\frac{5x + 3}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}
Step 2: Multiply both sides by the common denominator (x+1)(x−2)(x + 1)(x - 2) to clear the fractions.
5x+3=A(x−2)+B(x+1)5x + 3 = A(x - 2) + B(x + 1)
Step 3: Substitute x=2x = 2 to eliminate AA. This makes the (x−2)(x - 2) term vanish.
5(2)+3=A(0)+B(3)  ⟹  13=3B  ⟹  B=1335(2) + 3 = A(0) + B(3) \implies 13 = 3B \implies B = \frac{13}{3}
Step 4: Substitute x=−1x = -1 to eliminate BB. This makes the (x+1)(x + 1) term vanish.
5(−1)+3=A(−3)+B(0)  ⟹  −2=−3A  ⟹  A=235(-1) + 3 = A(-3) + B(0) \implies -2 = -3A \implies A = \frac{2}{3}
Step 5: Write the final decomposition using the values of AA and BB.
5x+3(x+1)(x−2)=2/3x+1+13/3x−2\frac{5x + 3}{(x + 1)(x - 2)} = \frac{2/3}{x + 1} + \frac{13/3}{x - 2}
Answer: 5x+3(x+1)(x−2)=23(x+1)+133(x−2)\frac{5x + 3}{(x + 1)(x - 2)} = \frac{2}{3(x + 1)} + \frac{13}{3(x - 2)}

Another Example

This example involves a repeated linear factor, which requires a different setup than distinct linear factors. You must include a separate term for each power of the repeated factor.

Problem: Decompose into partial fractions: 3x+5(x−1)2\frac{3x + 5}{(x - 1)^2}
Step 1: Since (x−1)(x - 1) is a repeated linear factor, the decomposition requires a term for each power of the factor, up to the highest power.
3x+5(x−1)2=Ax−1+B(x−1)2\frac{3x + 5}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}
Step 2: Multiply both sides by (x−1)2(x - 1)^2 to clear the denominators.
3x+5=A(x−1)+B3x + 5 = A(x - 1) + B
Step 3: Substitute x=1x = 1 to find BB directly.
3(1)+5=A(0)+B  ⟹  B=83(1) + 5 = A(0) + B \implies B = 8
Step 4: To find AA, compare coefficients of xx on both sides. The left side has coefficient 3 for xx, and the right side expands to Ax+(B−A)Ax + (B - A), so the coefficient of xx is AA.
A=3A = 3
Step 5: Write the final decomposition.
3x+5(x−1)2=3x−1+8(x−1)2\frac{3x + 5}{(x - 1)^2} = \frac{3}{x - 1} + \frac{8}{(x - 1)^2}
Answer: 3x+5(x−1)2=3x−1+8(x−1)2\frac{3x + 5}{(x - 1)^2} = \frac{3}{x - 1} + \frac{8}{(x - 1)^2}

Frequently Asked Questions

When do you use partial fractions?
You use partial fractions whenever you need to break a complicated rational expression into simpler pieces. The most common setting is in calculus, where partial fraction decomposition turns a difficult integral of a rational function into a sum of easier integrals (often involving natural logarithms or arctangent). It also appears in solving differential equations and working with Laplace transforms.
What is the difference between proper and improper rational expressions in partial fractions?
A proper rational expression has a numerator whose degree is strictly less than the degree of the denominator. Partial fraction decomposition applies directly only to proper rational expressions. If the expression is improper (numerator degree ≥ denominator degree), you must first perform polynomial long division to obtain a polynomial plus a proper rational remainder, and then decompose only the remainder.
How do you handle irreducible quadratic factors in partial fractions?
If the denominator contains an irreducible quadratic factor like (x2+bx+c)(x^2 + bx + c) that cannot be factored over the reals, the corresponding partial fraction term has the form Ax+Bx2+bx+c\frac{Ax + B}{x^2 + bx + c} — a linear numerator rather than just a constant. For a repeated irreducible quadratic raised to the nnth power, you include terms with linear numerators for each power from 1 up to nn.

Partial Fractions vs. Polynomial Long Division

Partial FractionsPolynomial Long Division
PurposeSplits a proper rational expression into simpler fractionsDivides a polynomial by another polynomial to get a quotient and remainder
When to useWhen the degree of the numerator is less than the degree of the denominatorWhen the degree of the numerator is greater than or equal to the degree of the denominator
Output formSum of simpler rational expressionsPolynomial quotient plus a proper rational remainder
RelationshipApplied after long division if the expression was originally improperOften a prerequisite step before partial fraction decomposition

Why It Matters

Partial fractions is one of the core techniques in a Calculus II integration course. Without it, you cannot integrate most rational functions, since expressions like 1x2−1\frac{1}{x^2 - 1} have no simple antiderivative in their combined form but split neatly into terms involving ln⁡∣x−1∣\ln|x - 1| and ln⁡∣x+1∣\ln|x + 1|. Beyond calculus, the method appears in engineering and physics when solving differential equations and computing inverse Laplace transforms.

Common Mistakes

Mistake: Using only one term for a repeated factor, e.g., writing A(x−1)2\frac{A}{(x-1)^2} instead of Ax−1+B(x−1)2\frac{A}{x-1} + \frac{B}{(x-1)^2}.
Correction: For a repeated linear factor (x−a)n(x - a)^n, you must include a separate fraction for every power from 1 through nn: A1x−a+A2(x−a)2+⋯+An(x−a)n\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}.
Mistake: Placing only a constant in the numerator over an irreducible quadratic factor, e.g., writing Ax2+4\frac{A}{x^2 + 4} instead of Ax+Bx2+4\frac{Ax + B}{x^2 + 4}.
Correction: An irreducible quadratic factor of degree 2 requires a linear expression Ax+BAx + B in its numerator. A single constant would leave you with too few unknowns to match the original expression.

Related Terms