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8.5 Partial Fractions (part 2)

last time: linear factors, some may be repeated

\[ \frac{1}{(x)(x-2)} = \frac{A}{x} + \frac{B}{x-2} \quad \text{find } A, B \]

\[ \frac{1}{(x)(x-2)^2} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{(x-2)^2} \quad \text{find } A, B, C \]

numerator must have lower degree than denominator

\[ \frac{26x^3 - 52x^2 + 2}{x^2 - 2x} \]

3rd degree numerator, 2nd degree denominator

must reduce before expansion

last time: rearrangement

this time: long division

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\[ \frac{26x^3 - 52x^2 + 2}{x^2 - 2x} \]

\[ = 26x + \frac{2}{x^2 - 2x} \]

Note: 2 is 0th degree, \(x^2-2x\) is 2nd degree

Polynomial Long Division

\(x^2 - 2x\) goes \(26x\) times into \(26x^3 - 52x^2 + 2\)

\[ \begin{array}{r} 26x \\ x^2 - 2x \enclose{longdiv}{26x^3 - 52x^2 + 2} \\ \underline{-(26x^3 - 52x^2) \phantom{+ 2}} \\ 2 \leftarrow \text{remainder} \end{array} \]

\[ = 26x + \frac{2}{x(x-2)} \]

Partial Fraction Decomposition of the remainder term:

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

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

Final Integral Setup:

\[ \int \left( 26x + \left( \frac{A}{x} + \frac{B}{x-2} \right) \right) dx \]

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Now Quadratic Factors

reducible: for example, \( x^2 - 2x = (x)(x - 2) \) product of linear factors
irreducible: cannot be factored into linear factors

for example, \( x^2 + 4 \)

\[ \frac{2x^2 - x + 4}{x^3 + 4x} = \frac{2x^2 - x + 4}{(x)(x^2 + 4)} \]

The denominator consists of a linear factor \( (x) \) and an irreducible quadratic factor \( (x^2 + 4) \).

\[ = \frac{A}{x} + \frac{Bx + C}{x^2 + 4} \]

For the linear factor \( x \):

Numerator is constant (0th degree)

\( 1^{\text{st}} \) degree denominator \( \rightarrow \) \( 0^{\text{th}} \) degree numerator

For irreducible quadratic:

Linear (\( 1^{\text{st}} \) deg) numerator

\( 2^{\text{nd}} \) degree denominator \( \rightarrow \) \( 1^{\text{st}} \) degree numerator

numerator is one degree lower than denominator

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Example

\[ \int \frac{x^2 + x + 2}{(x + 1)(x^2 + 1)} \, dx \]

degree check: numerator (2nd) < denom. (3rd) ok

Expansion:

\[ \frac{x^2 + x + 2}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \]

Find \( A, B, C \)

Note: \( (x+1) \) is linear, \( (x^2+1) \) is an irreducible quadratic.

Multiply by \( (x + 1)(x^2 + 1) \)

\[ x^2 + x + 2 = A(x^2 + 1) + (Bx + C)(x + 1) \]\[ = Ax^2 + A + Bx^2 + Bx + Cx + C \]\[ x^2 + x + 2 = (A + B)x^2 + (B + C)x + (A + C) \]

System of Equations (Equating Coefficients):

\( A + B = 1 \) — ①

\( B + C = 1 \) — ②

\( A + C = 2 \) — ③

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Solving for Partial Fraction Coefficients

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

(2): \( B + C = 1 \)

\( 1 - A + C = 1 \)

\( C = A \) sub into (3)

(3): \( A + C = 2 \)

\( A + A = 2 \rightarrow \)

A = 1

C = 1

B = 0

Integration using Partial Fractions

\[ \int \frac{x^2 + x + 2}{(x+1)(x^2+1)} dx = \int \left( \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \right) dx \]

\[ = \int \left( \frac{1}{x+1} + \frac{1}{x^2+1} \right) dx = \int \frac{1}{x+1} dx + \int \frac{1}{x^2+1} dx \]

trig sub \( \leftarrow \)

\[ = \ln |x+1| + \tan^{-1}(x) + D \]

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Repeated Irreducible Quadratic Factors

Repeated irreducible quadratic factors are handled just like how we handle repeated linear factors.

For example:

\[ \frac{1}{(x)(x^2+1)^2} = \frac{1}{(x)(x^2+1)(x^2+1)} \]

Note: \( x \) is linear; \( (x^2+1)(x^2+1) \) is irreducible and repeated.

\[ \frac{1}{(x)(x^2+1)(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \]

Find A, B, C, D, E

Multiply by \( (x)(x^2+1)(x^2+1) \)

\[ 1 = A(x^2+1)(x^2+1) + (Bx+C)(x)(x^2+1) + (Dx+E)(x) \]

Multiply out, collect by power:

\[ 1 = (A+B)x^4 + Cx^3 + (2A+B+D)x^2 + (C+E)x + A \]

\[ 0x^4 + 0x^3 + 0x^2 + 0x + 1 = (A+B)x^4 + Cx^3 + (2A+B+D)x^2 + (C+E)x + A \]

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Partial Fraction Decomposition: Solving for Coefficients

Right away we see:

\( C = 0 \)

\( A = 1 \)

Then:

\( A + B = 0 \rightarrow B = -A \rightarrow \)\( B = -1 \)

Then:

\( C + E = 0 \rightarrow E = -C \rightarrow \)\( E = 0 \)

\( 2A + B + D = 0 \)

\( 2(1) + (-1) + D = 0 \rightarrow \)\( D = -1 \)

Integration of the Rational Function

\[ \int \frac{1}{(x)(x^2+1)(x^2+1)} dx \]\[ = \int \left( \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \right) dx \]\[ = \int \left( \frac{1}{x} + \frac{-x}{x^2+1} + \frac{-x}{(x^2+1)^2} \right) dx = \int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx - \int \frac{x}{(x^2+1)^2} dx \]

easy

sub \( u = x^2 + 1 \) →

\[ \dots = \ln|x| - \frac{1}{2} \ln(x^2+1) + \frac{1}{2} \frac{1}{x^2+1} + F \]

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Partial Fraction Expansion Form

Write the form of expansion:

\[ \frac{2x^3 + 5x - 10}{x^2(x^2+4)^3(x^2-9)^2} = \frac{2x^3 + 5x - 10}{(x)(x)(x^2+4)(x^2+4)(x^2+4)(x+3)(x+3)(x-3)(x-3)} \]

\( (x+3)^2(x-3)^2 \)

\[ = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+4} + \frac{Ex+F}{(x^2+4)^2} + \frac{Gx+H}{(x^2+4)^3} + \frac{I}{x+3} + \frac{J}{(x+3)^2} + \frac{K}{x-3} + \frac{L}{(x-3)^2} \]