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### Partial Fractions

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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

x^2+7x+3        A       B       C
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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