i need some help with this partial fraction

Question

x²+4 ÷x² - 4

Answer 1

Since the degree of the numerator is not smaller than the degree of the denominator, we must first divide the numerator by the denominator.  

 

(x²+4)/(x²-4) = 1 + 8/[(x+2)(x-2)]

 

8/[(x+2)(x-2) = A/(x+2) + B/(x-2)

 

                 8 = A(x-2) + B(x+2)

 

          0x + 8 = (A+B)x + (-2A +2B)

 

            A+B = 0

         -2A + 2B = 8

 

Solving for A and B, we get A = -2 and B = 2.

 

The partial fraction decomposition for (x²+4)/(x²-4) is:  1 - 2/(x+2) + 2/(x-2)

 

Answer 2

I disagree. The answer is

 

[(x^2-2)/x]^2.

Check this out:

 

x^2 + 4/x^2 -4

=[(x^4 +4 -4x^2)/x^2]

=[(x^4 - 2*x^2*2 +2^2)/x^2]

=(x^2-2)^2/x^2)

=[(x^2-2)/x]^2.

Answer 3

Yes,  you need a way to get an x² into the numerator of the partial fraction so it's A/(x+2) + B/(x-2) + C.  To put the C over the common denominator, you'll multiply it by (x²-4) and get Cx²-4C, so C=1.  Then it reduces to a normal partial fraction with A = -B.  Interesting problem.

Answer 4

x^2 +4/ x^2 - 4 => x^2 + 4 / (x-2)(x+2), what can you do from here?