Given , by division we can reduce this as in the classical
case to . Now instead of complete factorization of
, we compute the * square-free factorization*:

Now we write

where each has degree less than (which is a much stronger requirement than being less than degree , of course). This is analogous to the familiar partial fraction decomposition where the repeated roots decomposed into .

Now let us consider integration of one of these terms, say
where **j > 1** (that is, we look at reducing the
repeated-root integrations to the simple root case).

Since is square-free, this means that . This
means that there exist polynomials **s** and **t** with degree **s < ** degree
degree , and degree **t < ** degree , such that

These polynomials **s** and **t** can be computed by the extended
Euclidean algorithm (see Geddes et al. for details).

Thus we may write our term to be integrated as

and after an integration by parts with **u = t** and , we
get

Notice that we have reduced the power of the denominator by 1. Carrying on recursively, we can show that

where

and

is square-free (i.e. has only simple roots). I was going to ask you,
in your homework, to fill in the details of a method based directly
on this formula (5), but the assignment is already large
enough. The basic idea is that you can write out the unknown coefficients
of and , differentiate the formula, and collect like terms
to get **m+n** equations in **m+n** unknowns. This avoids the sequence of
Hermite reductions and GCD calculations.

Thu Nov 16 13:46:20 PST 1995