Evaluate the integral. (Remember to use absolute values where appropriate. Use \( C \) for the constant of integration. \[ \int \frac{8}{(x-1)(x+7)} d x \]

To evaluate the integral \[ \int \frac{8}{(x-1)(x+7)} \, dx, \] we can use partial fraction decomposition. We first express the integrand as: \[ \frac{8}{(x-1)(x+7)} = \frac{A}{x-1} + \frac{B}{x+7}, \] where \( A \) and \( B \) are constants we need to solve for. Multiplying through by the denominator \( (x-1)(x+7) \) gives: \[ 8 = A(x+7) + B(x-1). \] Expanding the right-hand side, we have: \[ 8 = Ax + 7A + Bx - B = (A + B)x + (7A - B). \] To satisfy this equation for all \( x \), the coefficients of \( x \) must match and the constant terms must match: 1. \( A + B = 0 \) 2. \( 7A - B = 8 \) From the first equation, we can express \( B \) in terms of \( A \): \[ B = -A. \] Substituting \( B = -A \) into the second equation gives: \[ 7A - (-A) = 8 \implies 7A + A = 8 \implies 8A = 8 \implies A = 1. \] Now substituting \( A = 1 \) back to find \( B \): \[ B = -1. \] Thus, we have: \[ \frac{8}{(x-1)(x+7)} = \frac{1}{x-1} - \frac{1}{x+7}. \] Now we can integrate each term separately: \[ \int \frac{8}{(x-1)(x+7)} \, dx = \int \left( \frac{1}{x-1} - \frac{1}{x+7} \right) \, dx. \] This gives: \[ = \int \frac{1}{x-1} \, dx - \int \frac{1}{x+7} \, dx = \ln |x-1| - \ln |x+7| + C. \] Using the properties of logarithms, we can combine these terms: \[ = \ln \left| \frac{x-1}{x+7} \right| + C. \] Finally, the evaluated integral is: \[ \int \frac{8}{(x-1)(x+7)} \, dx = \ln \left| \frac{x-1}{x+7} \right| + C. \]