DE-BOOK Partial Fraction Decomposition

Section Partial Fraction Decomposition

[provisional cross-reference: WORK-IN-PROGRESS]

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. This process is especially useful when solving integrals and applying inverse Laplace transforms. The following steps outline the process to find the partial fraction decomposition of a rational function.

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Check the Degree of the Numerator and Denominator 🔗

Ensure the degree of the numerator is less than the degree of the denominator. If the numerator has a degree greater than or equal to the denominator, first perform polynomial long division to reduce the degree of the numerator.

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Factor the Denominator 🔗

Factor the denominator into irreducible linear or quadratic factors.

Linear factors: Expressions of the form \((s - a)\text{.}\)

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Irreducible quadratic factors: Expressions of the form \((s^2 + bs + c)\) where the discriminant \(b^2 - 4ac\) is negative.

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Set up the Partial Fraction Decomposition 🔗

Based on the factors of the denominator, write the decomposition:

For each linear factor \((s - a)\text{,}\) include a term of the form
\begin{equation*} \frac{A}{s - a}\text{.} \end{equation*}

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For repeated linear factors \((s - a)^n\text{,}\) include terms like:
\begin{equation*} \frac{A_1}{s - a} + \frac{A_2}{(s - a)^2} + \dots + \frac{A_n}{(s - a)^n}. \end{equation*}

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For irreducible quadratic factors \((s^2 + bs + c)\text{,}\) include a term of the form
\begin{equation*} \frac{Bs + C}{s^2 + bs + c}\text{.} \end{equation*}

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For repeated quadratic factors \((s^2 + bs + c)^n\text{,}\) include terms like:
\begin{equation*} \frac{B_1s + C_1}{s^2 + bs + c} + \frac{B_2s + C_2}{(s^2 + bs + c)^2} + \dots + \frac{B_ns + C_n}{(s^2 + bs + c)^n}. \end{equation*}

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Solve for the Constants 🔗

Multiply both sides of the equation by the common denominator and expand to eliminate fractions. Group terms by powers of \(s\) and set up a system of equations to solve for the unknown constants \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

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Final Result 🔗

Once the constants are found, write the final partial fraction decomposition. This decomposition can now be used for further calculations such as integrals or inverse Laplace transforms.

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✳️ Reference: Partial Fraction General Form Terms.

For each type of factor in the denominator, add the following terms to the partial fraction decomposition:

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Factor Type

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Factor

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Term(s) In General Form

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Linear

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\((s - a)\)

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\(\ds\frac{A}{s - a}\)

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Repeated Linear

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\((s - a)^n\)

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\(\ds\frac{A_1}{s - a} + \frac{A_2}{(s - a)^2} + \dots + \frac{A_n}{(s - a)^n}\)

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Irreducible Quadratic

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\((s^2 + bs + c)\)

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\(\ds\frac{Bs + C}{s^2 + bs + c}\)

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Repeated Quadratic

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\((s^2 + bs + c)^n\)

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\(\ds\frac{B_1s + C_1}{s^2 + bs + c} + \dots + \frac{B_ns + C_n}{(s^2 + bs + c)^n}\)

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🌌 Example 319.

Find the partial fraction decomposition form for each.

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\(\ds\frac{8x^2 - 51x + 41}{x^2 - 6x + 5}\)
Solution.
First, factor the denominator:

\begin{equation*} x^2 - 6x + 5 = (x - 5)(x - 1). \end{equation*}

The denominator has the following factors:

\(x-5\) (linear, single)

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\(x-1\) (linear, single)

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Hence, the FORM of the partial fraction decomposition is:

\begin{equation*} u(x) = \frac{A}{x-5} + \frac{B}{x-1}. \end{equation*}

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\(\ds\frac{17x - 11}{x^2(x-3)} \qquad\)
Solution.
Since the denominator is already factored, we see that the denominator has the following factors:

\(x\) (linear, double)

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\(x-3\) (linear, distinct)

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Hence, the FORM of the partial fraction decomposition is:

\begin{equation*} g(x) = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3}. \end{equation*}

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\(\ds\frac{x+12}{(x^2 - 4)^2} \qquad\)
Solution.
Here we need to finish factoring the denominator:

\begin{equation*} h(x) = \frac{x+12}{(x^2 - 4)^2} = \frac{x+12}{[(x-2)(x+2)]^2} = \frac{x+12}{(x-2)^2(x+2)^2} \end{equation*}

Now we see that the denominator has the following factors:

\((x-2)\) (linear, double)

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\((x+2)\) (linear, double)

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Hence, the FORM of the partial fraction decomposition is:

\begin{equation*} h(x) = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2} + \frac{D}{(x+2)^2}. \end{equation*}

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\(\ds\frac{3x-14}{(x+1)(x^2 + 2x + 1)} \qquad\)
Solution.

Here we need to finish factoring the denominator:

\begin{equation*} r(x) = \frac{3x-14}{(x+1)(x^2 + 2x + 1)} = \frac{3x-14}{(x+1)(x+1)(x+1)} = \frac{3x-14}{(x+1)^3} \end{equation*}

Now we see that \((x+1)\) (which is linear), is a factor (three times) of the denominator. Hence, the FORM of the partial fraction decomposition is:

\begin{equation*} r(x) = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{(x+1)^3}. \end{equation*}

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🌌 Example 320.

Find the partial fraction decomposition of

\begin{equation*} \frac{5s + 3}{(s - 1)(s^2 + s + 1)}\text{.} \end{equation*}

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Solution.

Factor the denominator as

\begin{equation*} (s - 1)(s^2 + s + 1)\text{.} \end{equation*}

The partial fraction decomposition is:

\begin{equation*} \frac{5s + 3}{(s - 1)(s^2 + s + 1)} = \frac{A}{s - 1} + \frac{Bs + C}{s^2 + s + 1}. \end{equation*}

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Multiply through by \((s - 1)(s^2 + s + 1)\) and solve for \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

\begin{equation*} 5s + 3 = A(s^2 + s + 1) + (Bs + C)(s - 1). \end{equation*}

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Expanding and comparing coefficients, we find:

\begin{equation*} A = 5, \quad B - A = 0, \quad C - B = 3. \end{equation*}

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Therefore, the partial fraction decomposition is:

\begin{equation*} \frac{5s + 3}{(s - 1)(s^2 + s + 1)} = \frac{5}{s - 1} + \frac{5s - 2}{s^2 + s + 1}. \end{equation*}

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Exercises Exercises

General Form.

Find the FORM of the partial fraction decomposition for each of the following. Make sure you completely factor each denominator before determining the decomposition. You need not find the values of the coefficients \(A,\) \(B,\) etc.

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