Multiplying Partial Fractions: Understanding the Rules

Jameseyboy
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Homework Statement





Homework 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!

The Attempt at a Solution

 

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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
 
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?
 
Jameseyboy said:
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?

Wolfram Alpha.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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