\( \int \frac{11}{(x-5)(x+6)} d x \) \( \int \frac{11}{(x-5)(x+6)} d x=\square \) (Use parentheses to clearly denote the argument of each function.)

To solve the integral \( \int \frac{11}{(x-5)(x+6)} \, dx \), we can use partial fraction decomposition. We express the integrand as: \[ \frac{11}{(x-5)(x+6)} = \frac{A}{x-5} + \frac{B}{x+6} \] Multiplying both sides by \((x-5)(x+6)\) to clear the denominator gives us: \[ 11 = A(x+6) + B(x-5) \] Expanding the right side: \[ 11 = Ax + 6A + Bx - 5B = (A + B)x + (6A - 5B) \] This leads to a system of equations: 1. \( A + B = 0 \) 2. \( 6A - 5B = 11 \) From the first equation, we get \( B = -A \). Substituting \( B \) into the second equation: \[ 6A - 5(-A) = 11 \implies 6A + 5A = 11 \implies 11A = 11 \implies A = 1 \] Then, substituting back to find \( B\): \[ B = -1 \] Now we can rewrite the integral: \[ \int \left( \frac{1}{x-5} - \frac{1}{x+6} \right) \, dx = \int \frac{1}{x-5} \, dx - \int \frac{1}{x+6} \, dx \] Integrating both terms: \[ \ln |x-5| - \ln |x+6| + C = \ln \left| \frac{x-5}{x+6} \right| + C \] So, the final answer is: \[ \int \frac{11}{(x-5)(x+6)} \, dx = \ln \left| \frac{x-5}{x+6} \right| + C \]