Integrating Reciprocals of Polynomials 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?
 A: 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.
A: 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.
