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Partial Fractions


Date: 06/08/97 at 15:18:37
From: Grahame Duke
Subject: Partial Fractions 

I am trying to revise for my maths exam next week, but I am having 
trouble with partial fractions.

How does 1/x^2*(x+2) resolve itself into A/2x^2 - 1/4x + 1/4(x+2)?

I would have thought that 1/x^2*(x+2) would turn into: 

A/x + B/x^2 + C/(x+2) 

I'm trying to solve a problem of integrating between 1 and 2 
(x^2+7x+3)/(x^2*(x+3))*dx and I am having trouble resolving the 
denominator into partial fractions. 

I think it should be x^2+7x+3 = A/x + B/x^2 + C/(x+3). That means that 
x^2+7x+3 = A(x)(x+3) + B(x+3) + C(x^2). (This is another process I 
have never fully grasped!) This means (by my counting) A = 7 (equate 
coefficients of x), B = 1, and C = -1.

Thus I would integrate 7lnx + 1/x - ln(x+3) 

Unfortunately, my calculator, which can handle integration of this 
type, doesn't agree with my answer - so I`ve slipped up somewhere.

Are there some quick and easy rules regarding partial fractions? My 
textbook only demonstrates easy problems.

Many thanks for any assistance, 
Grahame 


Date: 06/09/97 at 17:19:31
From: Doctor Anthony
Subject: Re: Partial Fractions 

Dear Grahame,

I will work through the problem, and you can then see where you have 
made mistakes.

             x^2+7x+3        A       B       C
Start with:  --------   =  ----- + ----- + ----- 
             x^2(x+3)        x      x^2    (x+3)  

Now if you multiply throughout by the denominator of the lefthand side 
you will get:

               x^2+7x+3 = Ax(x+3) + B(x+3) + Cx^2

There are two ways of finding A, B and C. Either multiply out the 
righthand side and arrange terms in x^2, x and a constant term. Then 
equate the coefficients of x^2 on each side, equate coefficients of x 
on each side and finally equate the constant term on both sides. This 
gives three equations from which you can find A, B and C.

A second method is to give x any convenient value, and this will give 
an equation connecting both sides. Give x another value and get a 
second equation, and then, if necessary, a third value to get a third 
equation. Often a mixture of the two methods is the quickest way to 
find A, B and C.

If we use x = 0, then the lefthand side equals 3 and the righthand 
side equals 3B, so B = 1.

Put in x = -3, then the lefthand side equals -9 and the righthand side 
equals 9C, so C = -1.

Finally, the coefficient of x^2 on the left is 1, and on the right 
A+C, so we can put:

                    A+C = 1   
                      A = 1-C
                      A = 1 - (-1)
                      A = 2

                                   2        1        1  
So the fraction can be written   ----- +  -----  - -----
                                   x       x^2     (x+3) 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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