Integrating Reciprocals of Polynomials

Question

I have seen integrals of the form $$\int \frac{1}{ax+b}dx$$ and $$\int \frac{1}{ax^{2}+bx+c} dx$$ But I cannot see how to integrate reciprocals of higher degree - does there exist a general solution to the integrals of reciprocals of cubics, quartics, and higher?

Partial Fractions. http://en.wikipedia.org/wiki/Partial_fraction — Weltschmerz, Jun 06 '12 at 12:27
I wonder how you propose to do, say, $\int(x^4+1)^{-1}\,dx$ by substitution. Surely not $u=x^2$, $du=2x\,dx$ leading to $\int(2\sqrt u(u^2+1))^{-1}\,du$? — Gerry Myerson, Jun 06 '12 at 13:07
Well spotted, I hadn't actually imagined making the substitution. I'll correct the question. — preferred_anon, Jun 06 '12 at 13:17
possible duplicate of: http://math.stackexchange.com/questions/20963/integration-by-partial-fractions-how-and-why-does-it-work — Eric Naslund, Jun 06 '12 at 16:41

score 7 · Accepted Answer · edited Apr 13 '17 at 12:19

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Reciprocals of higher degree can have their denominators factored into linear and/or irreducible quadratic terms, and from there, our result can be obtained through partial fraction decomposition.

For more details, see Arturo's excellent answer to this question.

edited Apr 13 '17 at 12:19

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answered Jun 06 '12 at 12:27

Cameron Buie

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Note that just because we know a partial-fraction decomposition exists doesn't mean that we can explicitly express the coefficients of that decomposition (any better than 'the $n$th root of the polynomial $p(x)$', at least); this is a consequence of Galois' famous result that polynomials of degree higher than 4 may have solutions not expressible through radicals. So we can say what form the integral will take, but not necessarily explicitly express the result 'elementarily'. – Steven Stadnicki Sep 18 '13 at 21:37
That's a good point, Steven. – Cameron Buie Sep 18 '13 at 23:05

score 1 · Answer 2 · answered Jun 06 '12 at 16:37

Assume the coefficients of the polynomial $f(x)$ are real: if they're not, the question will be answered quite differently.

Every root of the polyomial must then either be real or part of a pair of complex conjugates $a\pm bi$, where $a$ and $b$ are real.

Then $$ f(x) = c(x-\bullet)(x-\bullet)(x-\bullet)\cdots(x-\bullet) $$ where $c$ is the leading coefficient and each "$\bullet$" is one of the roots. If you get a real root, you've got a first-degree factor $(x-\bullet)$. If you get a pair of conjugates, then you have something like $$ (x - (3+5i)) (x - (3-5i)). $$ When you multiply this out, the imaginary parts cancel: $$ (x - (3+5i)) (x - (3-5i)) = x^2 - 3x - 5ix -3x + 5ix + 9 + 15i - 15i + 25 $$ $$ = 3x^2 - 6x + 34. $$ There you have a quadratic factor.

So you just get first-and second-degree factors.

(Finding just what those factors are, in the case of, e.g. a 15th-degree polynomial, can be quite a lot of work.)

How do we know that $f(x)$ factors as $$ c(x-\bullet)(x-\bullet)(x-\bullet)\cdots(x-\bullet)\ ? $$ That goes back to Carl Gauss in the year 1799. It is sometimes called the fundamental theorem of algebra, a name that some people object to on the grounds that it's a misnomer.

how can we for example factor x^4+1 into 1st and 2nd degree since it's factors are $((-1)^{(1/4)}-x) ((-1)^{(3/4)}-x) (x+(-1)^{(1/4)}) (x+(-1)^{(3/4)})$ — KFkf, Oct 08 '15 at 21:28
since when we multiply conjugates $((-1)^{(1/4)}-x)$ and $ (x+(-1)^{(1/4)})$ we get $-x^2+i$ — KFkf, Oct 08 '15 at 21:32

Integrating Reciprocals of Polynomials

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