\( \int \frac { x - 1 } { ( x + 1 ) ^ { 3 } } d x = \frac { A } { x + 1 } + \frac { B } { ( x + 1 ) ^ { 2 } } + \frac { C } { ( x + 1 ) ^ { 3 } } \)

To solve the integral \( \int \frac { x - 1 } { ( x + 1 ) ^ { 3 } } d x \) and determine the coefficients \( A, B, \) and \( C \) in the expression \( \frac { A } { x + 1 } + \frac { B } { ( x + 1 ) ^ { 2 } } + \frac { C } { ( x + 1 ) ^ { 3 } } \), we can use the method of partial fractions. First, we rewrite the integrand: \[ \frac{x - 1}{(x + 1)^3} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3} \] Next, we find a common denominator and equate the numerators: \[ x - 1 = A(x + 1)^2 + B(x + 1) + C \] Expanding the right-hand side gives: \[ x - 1 = A(x^2 + 2x + 1) + B(x + 1) + C = Ax^2 + (2A + B)x + (A + B + C) \] Now, we can compare coefficients on both sides: 1. **Coefficient of \( x^2 \)**: \( A = 0 \) 2. **Coefficient of \( x \)**: \( 2A + B = 1 \) → since \( A = 0 \), we get \( B = 1 \) 3. **Constant term**: \( A + B + C = -1 \) → substituting \( A = 0 \) and \( B = 1 \) gives \( 1 + C = -1 \) → thus \( C = -2 \) So we have \( A = 0 \), \( B = 1 \), and \( C = -2 \). Thus, the integral can be rewritten as: \[ \int \frac { x - 1 } { ( x + 1 ) ^ { 3 } } d x = \int \left( \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right) dx \] Now, integrate term by term: \[ = \left( -\frac{1}{x + 1} + \frac{2}{2(x + 1)^2} \right) + C = -\frac{1}{x + 1} + \frac{1}{(x + 1)^2} + C \] Therefore, the integral evaluates to: \[ -\frac{1}{x + 1} + \frac{1}{(x + 1)^2} + C \] Where \( C \) is the constant of integration.