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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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