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?
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5Partial Fractions. http://en.wikipedia.org/wiki/Partial_fraction – Weltschmerz Jun 06 '12 at 12:27
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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
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Well spotted, I hadn't actually imagined making the substitution. I'll correct the question. – preferred_anon Jun 06 '12 at 13:17
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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
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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.
Cameron Buie
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3Note 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
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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.
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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
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1since 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
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The general cubic cannot be done as single-summation series for except for certain cases, assuming a non-trivial factorization
For example it has the special values,
$$f(z)=\int_{0}^{1} \frac{-1}{x^3-2x^2+x-z}dx=\int_{0}^{1} \frac{1}{x^3-x^2+z}dx,,$$ which equals
$$ \sum_{m=0}^\infty \frac{(1)_{m} (1)_{2m}}{ z^{m+1} (2)_{3m}}. $$
For certain $z$ this factors into a rational linear term and a quadratic, so we obtain arctan
The reciprocal of a general cubic integral for some f(z) requires a double summation using the incomplete beta function integral
CarP24
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