Expert: Paul Klarreich - 9/25/2008

Question
Hi, i have read over and over the examples and answers using the partial fractions method, and i just dont get it!!
heres a question,

use the method of partial fractions to find an expression of the form shown for the function
f(x)=

   3x + 11   = A   +   Bx+C
(x-3) (x^2 +1)    x-3           x^2 + 1

(They are fractions e.g A/x-3 and Bx+C/x^2 + 1)

If you could do step by step answer using this method explaining each step, it would be such a huge help!!!

thankyou!


Answer
Questioner:   sarah
Category:  Calculus
Private:  No
 
Subject:  partial fractions
Question:  Hi, i have read over and over the examples and answers using the partial fractions method, and i just dont get it!!
heres a question,

use the method of partial fractions to find an expression of the form shown for the function
f(x)=

  3x + 11 = A + Bx+C
(x-3) (x^2 +1) x-3        x^2 + 1

(They are fractions e.g A/x-3 and Bx+C/x^2 + 1)

If you could do step by step answer using this method explaining each step, it would be such a huge help!!!

thankyou!
.....................................
Hi, Sarah,

Your problem should be written this way, and it will come through better:

 3x + 11             A      Bx + C
--------------- =  ------ + ---------
(x-3)(x^2 +1)       x - 3    x^2 + 1

You have the basics OK, that:

For a linear denominator like x-3, write a constant on top.
For a quadratic, write a linear top, like Bx + C.

So far so good.

NOW you must find a way to write three equations in A,B,C, that don't have x's.  
Start by multiplying out and clearing fractions:

 3x + 11             A      Bx + C
--------------- =  ------ + --------- << all times (x-3)(x^2+1)
(x-3)(x^2 +1)       x - 3    x^2 + 1

3x + 11 = A(x^2+1) + (Bx + C)(x - 3)

Now here is ONE way to proceed: First remove all parentheses:

3x + 11 = Ax^2 + A + Bx^2 - 3Bx + Cx - 3C

Now match up coefficients on both sides:

Const:  11 = A - 3C
x:       3 = -3B + C
x^2:     0 = A + B

Solve those equations for A,B,C and you are home.

Const:  11 = A - 3C
x:       3 = -3B + C
x^2:     0 = A + B  >>  A = - B, and plug that in:

Const:  11 = - B - 3C   >> times -3
x:       3 = -3B + C

Const: -33 = +3B + 9C
x:       3 = -3B + C
-------------------------  << add
     -30 = 10C
       C = -3

x:      3 = -3B - 3  << subst
       6 = -3B
       B = -2
       A = 2

Finish:

 3x + 11             2       2x + 3
--------------- =  ------  - ---------
(x-3)(x^2 +1)       x - 3     x^2 + 1

.........................................
Now here is ANOTHER way to proceed: Using

3x + 11 = A(x^2+1) + (Bx + C)(x - 3)

PICK three values of x.  x = 38, x = 19, x = -22 would work, but maybe

x = 3, x = 0, x = 1

would be a lot simpler.  That will produce three equations to be solved.  I'll leave that to you; it's a good exercise, as those authors like to say.