# Proving partial fraction decomposition? integral maybe?

1. Apr 10, 2006

### mr_coffee

hello everyone. I'm confused on what he wants here.
here are the directions:
http://img206.imageshack.us/img206/9692/lastscan2ju.jpg [Broken]

How do you prove such a thing? do i take the integral of the decompoisition and add them together or what? The homework and webworks never asked this type of question. It just was pratice on finding the decompoisition, but he has already given us that. Any help would be great!

I took the integral of the decompisition, the stuff on the right side of the equation and got:
8*ln(x-3)/9 + ln(x)/9 + 5/(3*(x-3))
and if i take the integral of the left hand side, it comes out to the same thing. is that all he wanted?

Last edited by a moderator: May 2, 2017
2. Apr 11, 2006

### Fermat

$$\frac{x^2-5x+1}{(x+3)^2x}=\frac{-5}{3(x-3)^2}+\frac{8}{9(x-3)}+\frac{1}{9x}$$

When doing partial fractions, all you do is write the lhs above as,

$$\frac{x^2-5x+1}{(x+3)^2x}=\frac{Ax+B}{(x-3)^2}+\frac{C}{(x-3)}+\frac{D}{x}$$

What you have is an identity, which means it is true for all values of x. So you can solve for A,B and C by putting in three different values of x to get three simultaneous eqns in A, B and C.

The preferred method of doing this is to multiply both sides by the denominator on the lhs. This will give you,

$$x^2-5x+1 = (Ax+B)x+C(x-3)x + D(x-3)^2$$

Now plug in x = 3 and x = 0, which are the zeros of the factors in the denominator of the (original) lhs.
You will still need to plug in some other value of x to get your third simultaneous eqn.

When dealing with factors on the bottom which are raised to some power, you do it like this,

$$\frac{1}{(x-c)^3} = \frac{(Ax^2 + Bx + c)}{(x-c)^3}+\frac{Dx+E}{(x-c)^2}+\frac{F}{(x-c)}$$

The term on the top should always be one power of x less than the term on the bottom.

Thus is your rational expression decomposed into partial fractions.

Doing this is the "proof" asked for.

3. Apr 11, 2006

### HallsofIvy

Staff Emeritus
One perfectly good method would be to do the addition indicated on the right: To add
$$\frac{-5}{3(x-3)^2}+\frac{8}{9(x-3)}+\frac{1}{9x}$$
you need a common denominator of 9x(x-3)2. Multiply the numerator and denominator of the first fraction by 3x, the second fraction by x(x-3), and the third fraction by (x-3)2 to get
$$\frac{-5(3x)}{9x(x-3)^2}+ \frac{8x(x-3)}{9x(x-3)^2}+ \frac{(x-3)^2}{9x(x-3)^2}$$
$$= \frac{-15x+ 8x^2- 24x+ x^2-6x+4}{9x(x-3)^2}$$
$$= \frac{9x^2- 45x+ 9}{9x(x-3)^2}$$
$$= \frac{9(x^2- 5x+ 1)}{9x(x-3)^2}$$
$$= \frac{x^2- 5x+ 1}{x(x-3)^2}$$

4. Apr 11, 2006

### mr_coffee

thanks guys for the help! that makes sense!