I need help solving:

integral of (20x-30) / (x(x-2)(x^2-2x+5))

If I try to decompose it;

I will have

LHS from above = (A/x)+(B/x-2) + ?

Stuck here.

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- Nov 20th 2009, 06:43 AMshadow85Partial fraction question
I need help solving:

integral of (20x-30) / (x(x-2)(x^2-2x+5))

If I try to decompose it;

I will have

LHS from above = (A/x)+(B/x-2) + ?

Stuck here.

- Nov 20th 2009, 06:50 AMlvleph
You can use the third factor:

$\displaystyle

\frac{20x-30}{x(x-2)(x^2-2x+5)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x^2 -2x+5}

$

C will a function of x. - Nov 20th 2009, 07:23 AMTWiX
$\displaystyle

\frac{20x-30}{x(x-2)(x^2-2x+5)} = \frac{A}{x} + \frac{B}{x-2} + \frac{Cx+D}{x^2 -2x+5}

$ - Nov 20th 2009, 07:45 AMshadow85
ohh I see, but I didnt think the polynomial could be there under Cx+D because its not factored?

So, now I am suppose to multiply both sides by (x(x-2)(x^2-2x+5))? - Nov 20th 2009, 12:04 PMTWiX
- Nov 20th 2009, 08:01 PMsimplependulum
$\displaystyle \frac{20x-30}{x(x-2)(x^2-2x+5)} $

$\displaystyle = \frac{20x-30}{(x^2-2x)((x^2-2x)+5)}$

If we let $\displaystyle u = x^2 - 2x $

and we know $\displaystyle \frac{1}{u(u+5)}= \frac{1}{5} \left ( \frac{1}{u} - \frac{1}{ u + 5 } \right ) $

The original fraction can be expressed in the form of :

$\displaystyle \frac{20x-30}{5} \left (\frac{1}{u} - \frac{1}{ u + 5 } \right ) $

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

we can see the last term $\displaystyle \frac{4x-6}{ x^2 -2x +5 } $

the numerator is a polynomial with degree $\displaystyle 1 $ , that means

we do not need to break it down into pieces anymore .

If we continue , we can obtain

$\displaystyle = \frac{2(2x-3)}{x(x-2)} - \frac{2(2x-3)}{ x^2 -2x +5 } $

$\displaystyle = \frac{ 2[(3/2)(x-2) + (1/2)x] }{x (x-2)} - \frac{2(2x-3)}{ x^2 -2x +5 } $

$\displaystyle = \frac{3}{x} + \frac{1}{x-2} - \frac{2(2x-3)}{x^2 -2x + 5}$ - Nov 21st 2009, 02:53 AMshadow85
umm wasnt the second post, enough to solve this?

- Nov 21st 2009, 07:58 AMlvleph
Not quite, because you to solve you would have to have a polynomial in the third fraction. I did say that C was a polynomial, but I didn't say of what order.