(A,B,C)

Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of 

\(\displaystyle \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}. \)
 \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}.

\(\small{ \begin{array}{|rcll|} \hline \dfrac{2(x^2+x-1)}{x(x^2-1)} &=& \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} \\ \\ \hline \\ \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} &=& \dfrac{A}{x} + \dfrac{B}{(x-1)} + \dfrac{C}{(x+1)} \qquad | \qquad \cdot x(x-1)(x+1) \\\\ 2(x^2+x-1) &=& \dfrac{A\cdot x(x-1)(x+1)}{x} + \dfrac{B\cdot x(x-1)(x+1)}{(x-1)} + \dfrac{C\cdot x(x-1)(x+1)}{(x+1)} \\\\ && \boxed{ 2(x^2+x-1) = A (x-1)(x+1) + B x (x+1) + C x(x-1) } \\\\ \hline \end{array} }\)

\(\small{ \begin{array}{lcll} \hline \mathbf{x = 0:} & 2(0+0-1) &=& A (0-1)(0+1) + B \cdot 0 (0+1) + C \cdot 0 (0-1) \\ & -2 &=& -A \\ & \mathbf{A} &\mathbf{=}& \mathbf{2} \\ \\ \hline \mathbf{x = 1:} & 2(1+1-1) &=& A (1-1)(1+1) + B\cdot 1 (1+1) + C \cdot 1(1-1) \\ & 2 &=& A \cdot 0(1+1) + B\cdot 1 (1+1) + C \cdot 1\cdot 0 \\ & 2 &=& 2B \\ & \mathbf{B} &\mathbf{=}& \mathbf{1} \\ \\ \hline \mathbf{x = -1:} & 2(1-1-1) &=& A (-1-1)(-1+1) + B\cdot (-1) (-1+1) + C \cdot (-1)(-1-1) \\ & -2 &=& A (-1-1)\cdot 0 + B\cdot (-1) \cdot 0 + C \cdot (-1)(-1-1) \\ & - 2 &=& 2C \\ & \mathbf{C} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array} }\)

\(\mathbf{(A,B,C) = (2,1,-1)}\)