Question

Is that correct?
Resolve into partial fraction\[\frac{x^2}{x^4-1}\]\[\frac{x^2}{x^4-1}=\frac{x^2}{(x^2+1)(x-1)(x+1)}\]
\[A(x-1)(x+1) + B(x^2+1)(x-1) + C(x^2+1)(x+1) = x^2\]\[(B+C) x^3+(A-B+C)x^2 +(B+C)x -A-B+C =x^2\]So, B+C = 0, A-B+C=1, A+B-C = 0
And I got A = 1/2, B=-1/4, C=1/4
So,
\[\frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} +\frac{1}{4(x-1)}\]