# Partial Fractions Explanation

1. Apr 10, 2014

### Jameseyboy

1. The problem statement, all variables and given/known data

2. Relevant equations

After looking through this on Wiki, I'm a little confused as to how these partial fractions are multiplied out. Is there a rule or something for this?

With simpler partials I can do it but this one is something else!

3. The attempt at a solution

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2. Apr 10, 2014

### HallsofIvy

Staff Emeritus
The denominator on the left is $$(x- 1)^3(x^2+ 1)^2$$ and every denominator on the right is a factor of that, so multiplying both sides by it will remove all fractions:

$$\frac{2x^6- 4x^5+ 5x^4- 3x^3+ x^2+ 3x}{(x- 1)^3(x^2+ 1)^2}((x-1)^3(x^2+ 1)^2)= 2x^6- 4x^5+ 5x^4- 3x^3+ x^2+ 3x$$

$$\frac{A}{x- 1}((x-1)^3(x^2+ 1)^2)= A(x- 1)^2(x^2+ 1)^2$$
$$\frac{B}{(x- 1)^2}((x-1)^3(x^2+ 1)^2)= B(x- 1)(x^2+ 1)^2$$
$$\frac{C}{(x- 1)^3}((x-1)^3(x^2+ 1)^2)= C(x^2+ 1)^2$$
$$\frac{Dx+ E}{x^2+ 1}((x-1)^3(x^2+ 1)^2)= (Dx+ E)(x- 1)^3(x^2+ 1)$$
$$\frac{Fx+ G}{(x^2+ 1)^2}((x-1)^3(x^2+ 1)^2)= (Fx+ G)(x- 1)^3$$

3. Apr 10, 2014

### Jameseyboy

Oh dear, had a complete 'idiot moment'. Thanks dude.

I've actually been googling trying to find an online differentiation calculator but I'm sure that some of them aren't giving the correct answers- do you know of any which are reliable?

4. Apr 10, 2014

### Ray Vickson

Wolfram Alpha.