Decompose \( \frac{x^{2}-3}{(x-1)\left(x^{2}+1\right)} \)

To decompose the given expression \( \frac{x^{2}-3}{(x-1)(x^{2}+1)} \), we can express it in the form: \[ \frac{A}{x-1} + \frac{Bx + C}{x^{2} + 1} \] where \( A \), \( B \), and \( C \) are constants to be determined. Multiply both sides by the denominator \( (x-1)(x^{2}+1) \): \[ x^{2}-3 = A(x^{2}+1) + (Bx + C)(x-1) \] Now, expand the right side: \[ x^{2}-3 = A(x^{2}+1) + (Bx^{2} - Bx + Cx - C) \] \[ = Ax^{2} + A + Bx^{2} + (C-B)x - C \] \[ = (A + B)x^{2} + (C-B)x + (A - C) \] Now we match coefficients from both sides of the equation: 1. For \( x^{2} \): \( A + B = 1 \) 2. For \( x \): \( C - B = 0 \) 3. Constant term: \( A - C = -3 \) Now we solve this system of equations: From \( C - B = 0 \), we have \( C = B \). Substituting \( C = B \) into \( A - C = -3 \): \[ A - B = -3 \quad \text{(1)} \] Now substituting \( C = B \) into \( A + B = 1 \): \[ A + B = 1 \quad \text{(2)} \] Now we can solve these two equations. From equation (2): \[ A = 1 - B \] Substituting this expression for \( A \) into equation (1): \[ 1 - B - B = -3 \] \[ 1 - 2B = -3 \] \[ -2B = -4 \] \[ B = 2 \] Using \( B = 2 \) in equation (2): \[ A + 2 = 1 \implies A = -1 \] And since \( C = B \): \[ C = 2 \] Thus, we have found the values: \[ A = -1, \quad B = 2, \quad C = 2. \] The partial fraction decomposition is: \[ \frac{x^{2}-3}{(x-1)(x^{2}+1)} = \frac{-1}{x-1} + \frac{2x + 2}{x^{2}+1}. \] So the final answer is: \[ \frac{-1}{x-1} + \frac{2x + 2}{x^{2}+1}. \]