# Integration with partial fractions

1. Jul 22, 2006

### Physics is Phun

stuck on this one question. mostly cause I don't know the proper steps for using partial fractions.

(4x-4)/(x^4 -2x^3 +4x^2 -6x +3)
which factors to
(4x-4)/(x^2+3)(x-1)^2
now I have the answer here. but I don't know the rules for decomposing this fraction. can someone go over them for me
this decomposes to A + B/(x-1)^2 + Cx+D/(x^2+3)
i don't understand this step, why is the A value by itself, and what is the determining factor that you have 2 terms in the numerator. I though it had something to do with the degree of the denominator. but then why is it B/(x-1)^2 for the one but Cx+D/(x^2+3) for the other. both are x^2 terms.

hope i've explained myself well enough,
thanks

2. Jul 22, 2006

### arildno

Well, first simplify your expression as follows:
$$\frac{4x-4}{(x^{2}+3)(x-1)^{2}}=\frac{4(x-1)}{(x^{2}+3)(x-1)^{2}}=\frac{4}{(x^{2}+3)(x-1)}$$

Now, assume an expansion of the form:
$$\frac{4}{(x^{2}+3)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+3}$$

3. Jul 22, 2006

### HallsofIvy

Staff Emeritus
The general "rules" (which I am sure are in your textbook) are these:
If a factor in the denominator is linear, i.e. (x-a), then you will need a fraction of the form $\frac{A}{x-a}$. That is, the numerator is a constant because the denominator is first degree.
If a factor in the denominator is a power of a linear term, i.e. (x-a)n, then you will need several fractions of the form $\frac{A}{x-a}$, $\frac{B}{(x-a)^2}$, up to the nth power: $\fra{Z}{(x-a)^n}$. Again, the denominator is always a constant because the "base" factor, x-a, is first power.
If a factor is quadratic and can't be factored, say ax2+ bx+ c, then you will need a fraction of the form $\frac{Ax+ B}{ax^2+ bx+ c}$. In each case, the numerator may be degree one less than the denominator (of course, A might be 0). If it were not, then you could divide it out.