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Contact Christian Bolander
Email christian.bolander@usu.edu

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Partial Fraction Expansion

Partial Fraction Expansion or Partial Fraction Decomposition is used to re-write a fraction. One of the possible reasons to do this is to make an expression much easier to integrate.

The Essentials

There are four cases of PFE used for different fractions depending on how factor-able the denominator is:

Case 1: Simple Factors

\frac{1}{(x - a) (x - b) (x - c)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}

Case 2: Repeated Factors

\frac{1}{(x - a)^{3} (x - b) (x - c)} = \frac{A}{x - a} + \frac{B}{(x - a)^{2}} + \frac{C}{(x - a)^{3}} + \frac{D}{x - b} + \frac{E}{x - c}

Case 3: Simple Irreducible Factors

\frac{1}{(x - a) (x^{2} + bx + c)} = \frac{A}{x - a} + \frac{B x + C}{x^{2} + bx + c}

Case 4: Repeated Irreducible Factors

\frac{1}{(a x^{2} + bx + c)^{3}} = \frac{A x + B}{a x^{2} + bx + c} + \frac{Cx + D}{(a x^{2} + bx + c)^{2}} + \frac{Ex + F}{(a x^{2} + bx + c^{3})}

Note that the fraction that you are re-writing doesn’t have to have just a one for the numerator, it can be any polynomial of order less than the denominator. Also, note that the number of coefficients (A, B, C, etc.) is always equal to the order of the denominator. After getting the fraction into one of these forms multiply the A, B, C, etc by the other terms to get two equal fractions with the same denominator:

\frac{1}{(x - a) (x - b) (x - c)} = \frac{A (x - b) (x - c) + B (x - a) (x - c) + C (x - a) (x - b)}{(x - a) (x - b) (x - c)}

1 = A (x - b) (x - c) + B (x - a) (x - c) + C (x - a) (x - b)

Then we set up a system of equations and solve for A, B, C, etc.

Example

Use partial fraction expansion to re-write the fraction:

\frac{x + 1}{x^{3} - 4 x^{2} + 4 x - 16}

\frac{x + 1}{(x - 4) (x^{2} + 4)}

\frac{x + 1}{(x - 4) (x^{2} + 4)} = \frac{A}{x - 4} + \frac{Bx + C}{x^{2} + 4}

\frac{x + 1}{(x - 4) (x^{2} + 4)} = \frac{A (x^{2} + 4)}{(x - 4) (x^{2} + 4)} + \frac{(Bx + C) (x - 4)}{(x^{2} + 4) (x - 4)}

\frac{x + 1}{(x - 4) (x^{2} + 4)} = \frac{A x^{2} + 4A + {Bx}^{2} - 4Bx + Cx - 4C}{(x - 4) (x^{2} + 4)}

x + 1 = A^{2} + 4A + {Bx}^{2} - 4Bx + Cx - 4C

(0) x^{2} + (1) x + 1 = (A + B) x^{2} + (C - 4 B) x + (4 A - 4 C)

{\begin{aligned} = 0 & A + B \\ = 1 & C - 4 B \\ = 1 & 4 A - 4 C \end{aligned}

{\begin{aligned} = \frac{1}{4} & A \\ = - \frac{1}{4} & B \\ = 0 & C \end{aligned}

\frac{x + 1}{x^{3} - 4 x^{2} + 4 x - 16} = \frac{\frac{1}{4}}{x - 4} + \frac{- \frac{x}{4}}{x^{2} + 4}

Practice

Use partial fraction expansion to re-write the fraction:

\frac{x + 1}{x^{5} + 6 x^{4} + 12 x^{3} + 8 x^{2}}

Solution:

\frac{- \frac{1}{16}}{x} + \frac{\frac{1}{8}}{x^{2}} + \frac{\frac{1}{16}}{x + 2} - \frac{\frac{1}{4}}{(x + 2)^{3}}

More Resources

Khan Academy: Intro to partial fraction expansion