Partial Fractions

1
Section 5.8

• Partial Fractions

2
Partial Fractions
3
Example 1

• Decompose into partial fractions
  
• Solution The degree of the numerator is less
  than the degree of the denominator.
• Begin by factoring the denominator (x 2)(2x ?
  3). We know that there are constants A and B such
  that
• 
  .
• To determine A and B, we add the expressions
  on the right

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Example 1 continued

• Equate the numerators 4x ? 13 A(2x ? 3) B(x
  2)
• Since the above equation containing A and B is
  true for all x, we can substitute any value of x
  and still have a true equation.
• If we choose x 3/2, then 2x ? 3 0 and A will
  be eliminated when we make the substitution.
• This gives us
• 4(3/2) ? 13 A2(3/2) ? 3 B(3/2 2)
• ?7 0 (7/2)B.
• B ?2.

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Example 1 continued

• If we choose x ?2, then x 2 0 and B will be
  eliminated when we make the substitution. So,
  4(?2) ? 13 A2(?2) ? 3 B(?2 2)
• ?21 ?7A.
• A 3.
• The decomposition is as follows
• 
  

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Example 2

• Decompose into partial fractions
• 
  
• Solution The degree of the numerator is 2
  and the degree of the denominator is 3, so the
  degree of the numerator is less than the degree
  of the denominator. The denominator is given in
  factored form. The decomposition has the
  following form
  

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Example 2 continued

• Next, we add the expression on the right
• 
  
• Then, we equate the numerators. This gives us
• 
  
• Since the equation containing A, B, and C is
  true for all of x, we can substitute any value of
  x and still have a true equation.
• In order to have 2x 1 0, we let x ½ .
  This gives us

Solving, we obtain A 5.
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Example 2 continued

• In order to have x ? 2 0, we let x 2.
• Substituting gives us
• To find B, we choose any value for x except ½ or
  2
• and replace A with 5 and C with ?2 . We let x
  1

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Example 2 continued

• The decomposition is as follows

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Example 3

• Decompose into partial fractions
  

11
Example 3 continued

• Solution
• The decomposition has
• the following form.
• Adding and equating numerators, we get
• or

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Example 3 continued

• We then equate corresponding coefficients
• 11 A 2C, The coefficients of the
  x2-terms
• ?8 ?3A B, The coefficients of the
  x-terms
• ?7 ?3B ? C. The constant terms
• We solve this system of three equations and
  obtain
• A 3, B 1, and C 4.
• The decomposition is as follows