Posted by Chris on Thursday, April 4, 2013 at 11:24am.
1) x^2
---------------
(x-1)(x^2+5x+4)
2) 3x^3-5x^2+12x+4
-----------------
x^4-16
3) 1
----------
x^2(x+1)^2
4) x+1
--------------
(x^2+1)(x-1)^2
I will do the first:
let
x^2/((x-1)(x+1)(x+5) = (A/(x-1) + B/(x+1) + C/(x+5)]/((x+1)(x-1)(x+5))
then obviously the common denominator is (x-1)(x+1)(x+5)
then only using the numerator:
A(x+1)(x+5) + B(x-1)(x+5) + C(x-1)(x+1) = x^2
let x = 1 --> 12A = 1 or A = 1/12
let x = 1 --> -8B=1 , or B = -1/8
let x = -5 --> 24C = 25 , or C = 24/25
then x^2/((x-1)(x^2 + 5x + 4))
= 1/(12(x-1)) - 1/((8(x+1)) + 25/(24(x+5))
3) is perhaps more interesting to do, as it's already factored you can just go ahead and solve this in the usual way. But there exist a much faster way to compute partial fraction expansions, which generalizes the method used by Reiny where you insert for x the critical values to simplify the equations.
When you have higher powers, you can in principle still use this method, by taking derivatives of both sides. But that's a bit cumbersome. What works even better is to expand the fraction around each critical point and keep only the singular terms. This is then the desired partial fraction expansion, because the difference between the function and the sum of all the expansions doesn't have any singularities anymore, it must therefore be a polynomial. However, it tends to zero at infinity, and must thus be equal to zero. This argument requires that you consider all the singularities jn the complex plane and that the numerator is of lower degree than the denominator.
The series expansion of
f(x) = 1/(x^2 (x+1)^2 )
around x = 0 is:
1/x^2 *(1 - 2 x + ...) =
1/x^2 - 2/x + non-singular terms.
The series expansion of
1/(x^2 (x+1)^2 )
around x = -1: Put x = -1+t:
1/t^2 *1/(1-t)^2 = 1/t^2 +2/t + non-singular terms =
1/(x+1)^2 + 2/(x+1) + non-singular terms.
The partial fraction expansion is thus:
1/x^2 - 2/x + 1/(x+1)^2 + 2/(x+1)