# Partial Fractions

1. Jul 14, 2011

### Deathfish

1. The problem statement, all variables and given/known data
Someone teach me how to write equations using the editor here...

This is part of an integration question

need partial fractions of 1/(x^2)(x-4)

2. Relevant equations

-nil-

3. The attempt at a solution

1/(x^2)(x-4)=(Ax+B)/(x^2)+C(x-4)

Put x=0 therefore B=-1/4
put x=4 therefore C= 1/16
Put x=1 therefore A=-1/16

1/(x^2)(x-4)=(-1/16)(x+4)/x^2 +1/16(x-4)

I get a different result on Mathematica, someone help

2. Jul 14, 2011

### micromass

Staff Emeritus
Hi Deathfish!

$$-\frac{x+4}{16x^2}+\frac{1}{16(x+4)}$$

then it is completely correct!!
Now, why does mathematica give another result? Well, the above result can be put in a form that is even more convenient. Indeed, the term $\frac{x+4}{16x^2}$ can be "simplified" further as

$$\frac{x+4}{16x^2}=\frac{1}{16x}+\frac{1}{4x^2}$$

This is a better form since it's more suitable for integration purposes. But your result isn't wrong!

3. Jul 14, 2011

### Staff: Mentor

The x2 factor in the denominator is not an irreducible quadratic, so there is no need for a term with Ax + b in the numerator.

I would decompose the expression in this way:
$$\frac{1}{x^2(x - 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 4}$$

How you did it is essentially the same as above, since Ax/x2 is the same as A/x.

4. Oct 25, 2011

### Dartx4

Could the A/X + B/X^2 also be written as Ax+B/X^2. I think you could because splitting it up into Ax/x^2 + B/x^2 is also seen as A/x + B/x^2. Also sorry about dead post reviving...very curious.

5. Oct 26, 2011

### Staff: Mentor

Yes, but you need parentheses. What you wrote would be considered to be Ax + (B/x2), and I'm certain that's not what you meant.

Write it as (Ax + B)/x2.