Can someone help me with this question?

1. Using the result , express in partial fractions.

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- July 13th 2008, 11:55 PMkleyzamPartial Fractions
Can someone help me with this question?

1. Using the result , express in partial fractions. - July 14th 2008, 12:01 AMMathstud28
- July 14th 2008, 12:02 AMkalagota
- July 14th 2008, 12:04 AMkalagota
- July 14th 2008, 12:14 AMkleyzam
I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:

- July 14th 2008, 12:15 AMMathstud28
- July 14th 2008, 12:18 AMkalagota
- July 14th 2008, 04:27 AMticbol
- July 14th 2008, 05:15 AMkalagota
- July 14th 2008, 05:25 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)]. - July 14th 2008, 05:29 AMkalagota
- July 14th 2008, 05:52 AMticbol
- July 14th 2008, 06:04 AMkalagota
well, given the situation he can always replace it by

, etc and add it to where it should be added.. :) - July 14th 2008, 06:19 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... - July 14th 2008, 06:27 AMkalagota