Partial fractions




Proper and improper rational fractions

If N(x) and D(x) are polynomials, then N(x)/D(x) is called a rational fraction.
If the degree of N(x) is less then the degree of D(x) the fraction is said to be a proper rational fraction, otherwise we call it an improper rational fraction.

Property of a improper rational fraction.

Each improper fraction can be written as the sum of a polynomial and a proper rational fraction. Indeed,
 
        N(x)                  rest(x)
        ---- = Quotient(x) + --------
        D(x)                   D(x)
Example
 
        3x2 + 4x                     7
        ---------- =  (3x + 7) + ---------
         x - 1                     x - 1

Partial fractions

Partial fractions are fractions of one of the following forms
 
           A               A
        -------   ;     ---------   ;
        (x - a)         (x - a)n


           Ax + B             Ax + B
        --------------  ;  -----------------    with b2 - 4c < 0
        x2 + bx + c        (x2 + bx + c)n


 Here A, B, b and c are real constants.
Examples
 
     5
  ---------
   (x -2)2

    2x + 3
  ------------
   x2 + x+ 4

    2x + 3
  ----------------
   (x2 + x+ 4)2

Partial fraction decomposition

Each proper rational fraction can be written as the sum of partial fractions.
Method:

Example 1

Given: the rational fraction
 
             2x + 5
        --------------------
        x3 - 3x2 - 4x + 12

  We factor the  denominator x3 - 3x2 - 4x + 12 = (x-3)(x+2)(x-2)

  Each factor causes exactly 1 elementary fraction of the sum


             2x + 5            A         B         C
        ---------------- =  ------- + ------- + -------
        (x-3)(x+2)(x-2)     (x - 3)   (x + 2)   (x - 2)

  A, B and C are presently unknown. Now we'll show how to calculate A, B and C.
  First we write the right side with one common denominator.
  Then the denominators  are equal on both sides.
  So, the numerators must be equal.

        (2x + 5) = A(x + 2)(x - 2) + B(x - 3)(x - 2) + C(x - 3)(x + 2)


        (2x + 5) = (A + B + C)x.x + (-5B - C)x + (-4A +6B -6C)

<=>
        / A + B + C = 0
        | -5B - C   = 2
        \ -4A +6B -6C=5

<=>
        . . .
<=>

        A = 2.2   B = 0.05   C = -2.25

Conclusion:
             2x + 5           2.2      0.05      2.25
        ---------------- =  ------- + ------- - -------
        (x-3)(x+2)(x-2)     (x - 3)   (x + 2)   (x - 2)

Example 2

Given: the rational fraction
 

                  7x
        ---------------------
        x3 - 6x2 + 12x - 8

  We factor the  denominator and we find  (x - 2)3
  This factor causes a sum of three partial fractions

           7x           A          B          C
        -------- =    -------  + -------  + -------
               3                       2          3
        (x - 2)      (x - 2)    (x - 2)    (x - 2)

  A, B and C are presently unknown. Now we'll show how to calculate A, B and C.
  First we write the right side with one common denominator.
  Then the denominators  are equal on both sides.
  So, the numerators must be equal.

        7x  = A (x - 2)2 + B (x-2) + C
<=>
        7x  = A x2 + (B - 4A) x + 4A - 2B + C

        with the same method as in previous example we calculate A, B and C.

        A = 0 ; B = 7; C = 14

Conclusion:

           7x            7          14
        -------- =    -------  + -------
               3             2          3
        (x - 2)       (x - 2)    (x - 2)

Example 3

Given: the rational fraction
 
            2                       2
          5x  + 4x + 1            5x  + 4x + 1
        ----------------- =     -----------------
          3    2        2           2        3
        (x  + x )(x + 1)          (x )(x + 1)

 The factor x2 in the denominator causes a sum of two partial fractions.
 The factor (x + 1)3 in the denominator causes a sum of three partial fractions.

            A     B        C          D          E
           --- + ---- +  -------  + -------  + -------
                   2                      2          3
            x     x     (x + 1)    (x + 1)    (x + 1)
  A, B, C, D and E are presently unknown. Now we'll show how to calculate A, B, C, D and E
  First we write the right side with one common denominator.
  Then the denominators  are equal on both sides.
  So, the numerators must be equal.


   5 x2  + 4x + 1 = A x (x + 1)3  + B (x + 1)3 + C x2 (x + 1)2 + D x2 (x + 1) + E x2

<=>

   5 x2  + 4x + 1 =(A + C) x4 + (3 A + B + 2C + D)x3 + (3A + 3B + C + D + E)x2 + (A + 3B)x + B

<=>

 /  A      +  C          = 0
 | 3 A + B + 2C + D      = 0
 | 3 A + 3B + C + D + E  = 5
 |  A + 3B               = 4
 \      B                = 1

<=>
   ...
<=>
        A = 1 ; B = 1; C = -1 ; D = -2 ; E = 2

Conclusion:

            1     1        -1         -2        2
           --- + ---- +  -------  + -------  + -------
                   2                      2          3
            x     x     (x + 1)    (x + 1)    (x + 1)

Example 4

Given: the rational fraction
 

                (2x + 1)
        ------------------------
           2        2
        ( x  + 1)( x  +  x + 1)

  The denominator can't be factored in real factors of lower degree.
  Each factor in the denominator gives us just one elementary fraction  with a numerator
  of the first degree.

                (2x + 1)              Ax + B            Cx + D
        ----------------------  =  --------------- + -----------------
           2        2                   2                2
        ( x  + 1)( x  +  x + 1)      ( x  + 1)        ( x  +  x + 1)

  A, B, C and  D  are presently unknown. Now we'll show how to calculate A, B, C, and D.
  First we write the right side with one common denominator.
  Then the denominators  are equal on both sides.
  So, the numerators must be equal.


       (2x + 1) = (Ax + B)(x2 + x + 1) + (Cx + D)(x2 + 1)

<=>

       (2x + 1) = (A + C) x3 + (A + B + D) x2 + (A + B + C) x + B + D

       and with the aid of a system with 4 unknowns, we find
        ...

        A = -1 ; B = 2 ; C = 1; D = -1
Conclusion



                (2x + 1)              - x + 2            x - 1
        ----------------------  =  --------------- + -----------------
           2        2                   2                2
        ( x  + 1)( x  +  x + 1)      ( x  + 1)        ( x  +  x + 1)

Example 5

Given: the rational fraction
 
      x4 + 5x3 + 16x2 + 26x + 22
     ------------------------------
         x3 + 3 x2 + 7x + 5

   This improper fraction can be written as the sum of a polynomial and a
    proper rational fraction.

                 3 x2 + 7 x + 12
     (x+2) + -------------------------
              x3 + 3 x2 + 7x + 5

  We factor the denominator in (x+1)(x2 + 2x + 5)

   3 x2 + 7 x + 12             A            B x + C
  ---------------------- = ------------ +  -----------------
   (x+1)(x2 + 2x + 5)       x + 1          x2 + 2x + 5

  A, B and C are presently unknown.
  First we write the right side with one common denominator.
  Then the denominators  are equal on both sides.
  So, the numerators must be equal.

  3x2 + 7x + 12 = A(x2 + 2x + 5) + (Bx + C)(x + 1)

  As above, you can find A, B and C by solving a system of three equations.

  A = 2 ; B = 1 ; C = 2

  3 x2 + 7 x + 12             2              x + 2
  ---------------------- = ------------ +  -----------------
   (x+1)(x2 + 2x + 5)       x + 1          x2 + 2x + 5

Exercises

Write the left side as sum of partial fractions and verify your answer.
 
           16                  1                1                4
    ------------------   =  -----------   -  -----------   -  ------------
    x3 - x2 - 5x - 3         (x - 3)          (x + 1)          (x + 1)2


   - 3 x3 + 8 x2 - 4 x + 5                x            3         1
  ---------------------------------- = ------------ + ------- - ------
   - x4 + 3 x3  - 3 x2 + 3 x - 2         x2 + 1       x - 1     x - 2



   2 x3 + 7 x2 - 2 x + 6            2x                  3
   ------------------------ = ------------------ + ---------------
        x4 + 4                  x2  - 2 x + 2         x2  + 2 x + 2






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