by **HallsofIvy** » Fri Jan 15, 2021 11:31 am

You need "partial fractions". [tex]x^4- 16= (x^2- 4)(x^2+ 4)= (x- 2)(x+ 2)(x^2+ 4)[/tex].

[tex](x^2- 16)^2= (x- 2)^2(x+ 2)^2(x^2+ 4)^2[/tex].

That can be expanded in partial fractions as

[tex]\frac{A}{x- 2}+ \frac{B}{(x- 2)^2}+ \frac{C}{x+ 2}+ \frac{D}{(x+ 2)^2}+ \frac{Ex+ F}{x^2+ 4}+ \frac{Gx+ H}{(x^2+ 4}[/tex].

You will need to solve for A, B, C, D, E, F, G, and H.

With [tex]\frac{1}{(x^4- 16)^2}= \frac{A}{x- 2}+ \frac{B}{(x- 2)^2}+ \frac{C}{x+ 2}+ \frac{D}{(x+ 2)^2}+ \frac{Ex+ F}{x^2+ 4}+ \frac{Gx+ H}{(x^2+ 4}[/tex] multiply on both sides by [tex](x^4- 16)^2[/tex]. On the left we have, of course, 1. On the right one or two of the various factors will cancel leaving

[tex]1= A(x-2)(x+ 2)^2(x^2+ 4)^2+ B(x+ 2)^2(x^2+ 4)^2+ C(x- 2)^2(x+ 2)(x^2+ 4)^2+ D(x- 2)^2(x^2+ 4)^2+ (Ex+ F)(x- 2)^2(x+ 2)^2(x^2+ 4)+ (Gx+ H)(x- 2)^2(x+ 2)^2[/tex].

That equation is to be true for all x. Taking 8 values for x will give us 8 equations to solve for A, B, C, D, E, F, G, and H. In particular, if we take x= 2, since every coefficient except that of B has a factor of x- 2, we have simply [tex]1= B(4^2)(8^2)= 1024B[/tex] so [tex]B= \frac{1}{1024}[/tex]. Similarly, taking x= -2, we have simply

[tex]1= D(-4)^2(8^2)= 1024D[/tex] so D is also [tex]\frac{1}{1024}[/tex].

The others can' t be reduced so simply but replacing x with 6 different numbers (other than 2 or -2) gives 6 equations to solve for A, C, E, F, G, and H.