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8.5 Partial Fractions Expansions (part 1)

\[ \int \frac{x-8}{x^2-7x+10} dx \]

options:

direct integration: no.
substitution (u): no
by parts: maybe, but messy
trig subs: maybe, but messy

if we could recognize that

\[ \frac{x-8}{x^2-7x+10} = \frac{2}{x-2} - \frac{1}{x-5} \]

then

\[ \begin{aligned} \int \frac{x-8}{x^2-7x+10} dx &= \int \left( \frac{2}{x-2} - \frac{1}{x-5} \right) dx \\ &= \int \frac{2}{x-2} dx - \int \frac{1}{x-5} dx \end{aligned} \]

\[ = 2 \ln |x-2| - \ln |x-5| + C \]

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How do we do partial fractions expansion?

Case 1: Denominator is a product of distinct linear factors

not repeatedfirst order (power is 1)

\[ \frac{x-8}{x^2-7x+10} = \frac{x-8}{(x-5)(x-2)} = \frac{A}{x-5} + \frac{B}{x-2} \]

A, B : constants

now find A and B

\[ \frac{x-8}{(x-5)(x-2)} = \frac{A}{x-5} + \frac{B}{(x-2)} \]

multiply through by \( (x-5)(x-2) \)

\[ \begin{aligned} x-8 &= A(x-2) + B(x-5) \\ &= Ax - 2A + Bx - 5B \\ x-8 &= (A+B)x + (-2A-5B) \end{aligned} \]

Equating coefficients of like terms:

\[ A+B = 1 \quad \text{--- (1)} \]\[ -2A-5B = -8 \quad \text{--- (2)} \]

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From (1) → \( B = 1 - A \) sub into (2)

\[ -2A - 5B = -8 \]\[ -2A - 5(1 - A) = -8 \]\[ -2A - 5 + 5A = -8 \]\[ 3A = -3 \quad \rightarrow \quad A = -1 \]\[ B = 1 - A = 1 - (-1) = 2 \]

So,

\[ \frac{x - 8}{x^2 - 7x + 10} = \frac{A}{x - 5} + \frac{B}{x - 2} = -\frac{1}{x - 5} + \frac{2}{x - 2} \]

Same as shown on first page

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Example

\[ \int \frac{1}{x^2 - 4} dx \]

\[ \frac{1}{x^2 - 4} = \frac{1}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2} \]

product of distinct linear factors

multiply by \( (x - 2)(x + 2) \)

\[ 1 = A(x + 2) + B(x - 2) \]\[ = Ax + 2A + Bx - 2B \]\[ 0x + 1 = (A + B)x + (2A - 2B) \]

\( 0x + 1 = (A + B)x + (2A - 2B) \)
Equating coefficients of like terms

\[ A + B = 0 \quad \text{--- (1)} \]\[ 2A - 2B = 1 \quad \text{--- (2)} \]

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Another way to solve the system

multiply (1) by 2

\[\begin{aligned} 2A + 2B &= 0 \quad - (3) \\ 2A - 2B &= 1 \quad - (2) \end{aligned}\]

add (3) and (2) \(\rightarrow 4A = 1 \rightarrow A = \frac{1}{4}, \quad B = -\frac{1}{4}\) (from any eq., for example (1))

\[\begin{aligned} \int \frac{1}{x^2 - 4} dx &= \int \left( \frac{A}{x-2} + \frac{B}{x+2} \right) dx \\ &= \int \frac{1}{4} \frac{1}{x-2} dx + \int -\frac{1}{4} \frac{1}{x+2} dx \end{aligned}\]

\[ = \frac{1}{4} \ln |x-2| - \frac{1}{4} \ln |x+2| + C \]

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Case 2: Denominator is a product of linear factors and some of which are repeated

\[ \frac{10}{5x^2 - 2x^3} = \frac{10}{(x^2)(5-2x)} = \frac{10}{(x)(x)(5-2x)} \]

repeated linear factor

for repeated factors

\[ \frac{10}{(x)(x)(5-2x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{5-2x} \]

each time the factor appears, bump power up by 1

multiply by \((x)(x)(5-2x)\)

\[ 10 = A(x)(5-2x) + B(5-2x) + C(x)(x) \]

another way to find constants: pick x such that as many of them as possible disappear

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Partial Fraction Decomposition: Solving for Constants

Pick \( x = 0 \)

\[ 10 = B(5 - 0) \rightarrow 10 = 5B \rightarrow B = 2 \]

Pick \( x = \frac{5}{2} \) (such that \( 5 - 2x = 0 \))

\[ 10 = C\left(\frac{5}{2}\right)\left(\frac{5}{2}\right) \rightarrow 10 = \frac{25}{4}C \rightarrow C = \frac{8}{5} \]

For the last constant, choose any convenient \( x \) and use the constants we already know.

Pick \( x = 1 \)

\[ 10 = 3A + 3B + C \]

Substituting \( B = 2 \) and \( C = \frac{8}{5} \):

\[ = 3A + 6 + \frac{8}{5} \rightarrow A = \frac{4}{5} \]

Final Decomposition

\[ \frac{10}{5x^2 - 2x^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{5-2x} = \frac{4}{5} \cdot \frac{1}{x} + \frac{2}{x^2} + \frac{8}{5} \cdot \frac{1}{5-2x} \]

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Integration using Partial Fractions

\[ \int \frac{10}{5x^2 - 2x^3} dx = \int \left( \frac{4/5}{x} + \frac{2}{x^2} + \frac{8/5}{5-2x} \right) dx \]

\[ = \dots = \frac{4}{5} \ln |x| - \frac{2}{x} - \frac{4}{5} \ln |5 - 2x| + D \]

Repeated Linear Factors Example

If we have:

\[ \frac{1}{(x)(x)(x-1)(x-1)(x-1)(x+10)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3} + \frac{F}{x+10} \]

Note: The factor \( x \) appears twice, and the factor \( (x-1) \) appears thrice.

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Partial Fraction Expansion Requirements

Important: for the expansion to work, the numerator of the original expression MUST have a lower degree than the denominator.

Example 1: Valid Expression

\[ \frac{x - 8}{x^2 - 7x + 10} \]

← first degree (1)← second degree (2)

is ok.

Example 2: Invalid Expression

\[ \frac{-2x^3 + 5x^2 + 10}{-2x^3 + 5x^2} \]

← 3rd← 3rd

is not ok

BEFORE expansion, we need to do:

\[ \frac{-2x^3 + 5x^2}{-2x^3 + 5x^2} + \frac{10}{-2x^3 + 5x^2} = 1 + \frac{10}{-2x^3 + 5x^2} \]

can expand as we did w/ previous examples