# Integration of rational functions

1. Mar 6, 2004

### Kuja

How do I solve the integration of a rational function such as:

x^2 - 6x - 2
(x^2 + 2)^2

If possible, please list the general rule of solving, I DO NOT want the answer, I simply want to know the way of solving it.

2. Mar 7, 2004

### HallsofIvy

Staff Emeritus
The "rule" that you want is "partial fractions".

If you have a rational function, in which the denominator can be factored into distinct linear factors, such as
(x- 3)/((x+1)(x-2)), then you can write it as a sum of fractions, each having one factor as denominator:
(x-3)/((x+1)(x-2))= A/(x+1)+ B/(x-2).
(Of course, the numerator is of lower degree than the denominator: if not, divide first.)

If some of the linear factors are repeated, such as
(x+ 4)/((x+1)2(x-2)), then you will need all powers of that repeated factor: A/(x+1)+ B/(x+1)2+ C/(x-2)

If some of the factors are quadratics that cannot be factored, then they can, by completing the square, be written in the form "a(x-b)2+ c" and you will need a fraction of the form (Ax+ B)/(a(x-b)2+c), for example (3x2- 2x+ 4)/((x2+ 4)(x+3)) can be written (Ax+B)/(x2+4)+ C/(x+3).

In this particular example,
$$\frac{x^2-6x- 2}{(x^2+2)^2)^2}$$
can be written in the form
$$\frac{Ax+B}{(x^2+2)^2}+\frac{Cx+D}{x^2+2}$$

Those have to be equal for all x so one way of finding A, B, C, D is by setting those equal:
$$\frac{x^2-6x- 2}{(x^2+2)^2)^2}= \frac{Ax+B}{(x^2+2)^2}+\frac{Cx+D}{x^2+2}$$
Now multiply both sides by that denominator to clear the fractions and set x equal to 4 different numbers to get 4 equations for A, B, C, and D. You can often choose those numbers to simplify the equations.