The integral in this exercise converges. Evaluate the integral without using a table. \( \int_{-\infty}^{-6} \frac{50}{x^{2}-25} d x \) \( \int_{-\infty}^{-6} \frac{50}{x^{2}-25} d x=\square \) (Type an exact answer.)

To evaluate the integral \( \int_{-\infty}^{-6} \frac{50}{x^{2}-25} d x \), we can start by simplifying the integrand: The integrand can be rewritten using partial fraction decomposition: \[ \frac{50}{x^{2}-25} = \frac{50}{(x-5)(x+5)}. \] Now, we can set it up for partial fraction decomposition: \[ \frac{50}{(x-5)(x+5)} = \frac{A}{x-5} + \frac{B}{x+5}. \] Multiplying through by \( (x-5)(x+5) \) gives: \[ 50 = A(x+5) + B(x-5). \] Expanding this, we have: \[ 50 = Ax + 5A + Bx - 5B = (A + B)x + (5A - 5B). \] Setting coefficients equal, we have: 1. \( A + B = 0 \) (1) 2. \( 5A - 5B = 50 \) (2) From equation (1), we find \( B = -A \). Substituting this into equation (2) gives: \[ 5A - 5(-A) = 50 \implies 10A = 50 \implies A = 5 \quad \text{and} \quad B = -5. \] Thus, we can rewrite the integrand: \[ \frac{50}{(x-5)(x+5)} = \frac{5}{x-5} - \frac{5}{x+5}. \] Now we can integrate: \[ \int \left( \frac{5}{x-5} - \frac{5}{x+5} \right) dx = 5 \ln |x-5| - 5 \ln |x+5| + C = 5 \ln \left| \frac{x-5}{x+5} \right| + C. \] Now we compute the improper integral: \[ \int_{-\infty}^{-6} \left( \frac{5}{x-5} - \frac{5}{x+5} \right) dx. \] As \( x \to -\infty \): 1. \( 5 \ln |x-5| \to 5 \ln (-x) \) 2. \( 5 \ln |x+5| \to 5 \ln (-x) \) So as \( x \to -\infty \), the difference approaches \( 5 \ln \left| \frac{-x-5}{-x+5} \right| \). Calculating as \( x \) approaches -6: \[ \lim_{x \to -6} \left( 5 \ln \left| \frac{x-5}{x+5} \right| \right) = 5 \ln \left| \frac{-6-5}{-6+5} \right| = 5 \ln \left| \frac{-11}{-1} \right| = 5 \ln 11. \] Thus, the value of the integral is: \[ \int_{-\infty}^{-6} \frac{50}{x^{2}-25} d x = 5 \ln 11. \] So we find \( \int_{-\infty}^{-6} \frac{50}{x^{2}-25} d x = 5 \ln 11 \).