<s>[INST] Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions How does one write 
$$\frac{x^4(1-x)^4}{1+x^2}$$
in terms of partial fractions? 
My Attempt
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$
$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$
Multiplying out and comparing coefficient gives:
$$=x^6-4x^5+5x^4-4x^2+0+\frac{0}{1+x^2}$$
This is obviously not correct, how to do this properly?
 [/INST] You should have tried
$$\frac{x^4(1-x)^4}{1+x^2}=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx +\mathbf{J}+\frac{\mathbf{G + Hx}}{1+x^2}$$
By the way, multiplying out and comparing coefficients gives:
$$\frac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+\mathbf{4}+\frac{\mathbf{-4}}{1+x^2}$$
</s>