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October 24th, 2018, 04:54 AM   #11
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Quote:
 Originally Posted by SDK To expand on what skipjack has done, EVERY polynomial of degree $d > 2$ can be factored into lower degree polynomials over the reals. Also EVERY polynomial of degree $d > 1$ can be factored into lower degree polynomials over the complex numbers. Hence, partial fractions works universally for problems like this.
Yes, in theory a problem like this is always solvable with partial fractions, but consider this:

$$\int \frac{1 - 5x^4}{(x^5 - x - 1)^2}dx$$

In theory it is indeed solvable with partial fractions. Maybe in practice too, I haven't tried it. But it's going to be very complicated because you can't find the explicit roots of the denominator. So I'm not at all sure if the method will work here.

However, an easy solution does exist here, a primitive is simply
$$\frac{1}{x^5 - x - 1}.$$

October 25th, 2018, 11:02 AM   #12
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Quote:
 Originally Posted by SDK To expand on what skipjack has done, EVERY polynomial of degree $d > 2$ can be factored into lower degree polynomials over the reals. Also EVERY polynomial of degree $d > 1$ can be factored into lower degree polynomials over the complex numbers. Hence, partial fractions works universally for problems like this.
True but the method of assuming Ax+B in the numerator works for only those polynomials in the denominator which have real rational factors. This I think is true..If its wrong, plz correct me. Thx.

October 25th, 2018, 11:04 AM   #13
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Quote:
 Originally Posted by romsek If I understand the problem... what you need to do is set up $\dfrac{x^3-1}{(x^4+1)(x+1)} = \dfrac{A x^3 + B x^3 + C x + D}{x^4+1}+\dfrac{E}{x+1}$
Oh I see...Actually I wanted to know the numerator which is cubic with constants A, B, C & D. Thanks. This is to be assumed.

 Tags integral, plz, solve

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