Can someone help me with this question?

1. Using the result $\displaystyle x^3 + 1 = (x+1)(x^2 - x + 1)$, express $\displaystyle \frac{x^5 - 1}{x^2(x^3 + 1)}$ in partial fractions.

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- Jul 13th 2008, 10:55 PMkleyzamPartial Fractions
Can someone help me with this question?

1. Using the result $\displaystyle x^3 + 1 = (x+1)(x^2 - x + 1)$, express $\displaystyle \frac{x^5 - 1}{x^2(x^3 + 1)}$ in partial fractions. - Jul 13th 2008, 11:01 PMMathstud28
- Jul 13th 2008, 11:02 PMkalagota
- Jul 13th 2008, 11:04 PMkalagota
- Jul 13th 2008, 11:14 PMkleyzam
I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:

$\displaystyle -x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1)$ $\displaystyle + Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)$ - Jul 13th 2008, 11:15 PMMathstud28
- Jul 13th 2008, 11:18 PMkalagota
- Jul 14th 2008, 03:27 AMticbol
- Jul 14th 2008, 04:15 AMkalagota
- Jul 14th 2008, 04:25 AMticbol
He divided, or simplified, the original LHS, but he did not do that in the original RHS?

Then his final "equation or identity", even if the last term in the new RHS is corrected, is wrong.

Because if he multiplied the original RHS by (x^2)(x^3 +1), then he should have multiplied the original LHS by the same amount.

Clearly, (-x^2 -1) is not [(x^5 -1) / (x^2)(x^3 +1)] * [(x^2)(x^3 +1)]. - Jul 14th 2008, 04:29 AMkalagota
- Jul 14th 2008, 04:52 AMticbol
- Jul 14th 2008, 05:04 AMkalagota
well, given the situation he can always replace it by

$\displaystyle \frac{x}{x}, \frac{x^2}{x^2}$, etc and add it to where it should be added.. :) - Jul 14th 2008, 05:19 AMticbol
You lost me there.

So the final decomposition might be:

x/x + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?

or, (x^2 / x^2) + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?

Etc....

And so the the final integral would be

= x +{INT[the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]dx} +C?

Umm... - Jul 14th 2008, 05:27 AMkalagota