1. We are asked to determine if the improper integral $$\int_1^\infty \frac{1}{x-6} \, dx$$ converges or diverges, and if it converges, to find its value. 2. The integral is improper because the upper limit is infinity and the integrand has a vertical asymptote at $$x=6$$ (since the denominator becomes zero there). 3. We split the integral at the point of discontinuity $$x=6$$: $$\int_1^\infty \frac{1}{x-6} \, dx = \int_1^6 \frac{1}{x-6} \, dx + \int_6^\infty \frac{1}{x-6} \, dx$$ 4. Evaluate the first integral as a limit approaching 6 from the left: $$\int_1^6 \frac{1}{x-6} \, dx = \lim_{t \to 6^-} \int_1^t \frac{1}{x-6} \, dx$$ 5. The antiderivative of $$\frac{1}{x-6}$$ is $$\ln|x-6|$$, so: $$\int_1^t \frac{1}{x-6} \, dx = \left[ \ln|x-6| \right]_1^t = \ln|t-6| - \ln|1-6| = \ln|t-6| - \ln 5$$ 6. Taking the limit as $$t \to 6^-$$, note that $$t-6 \to 0^-$$, so $$|t-6| \to 0$$ and $$\ln|t-6| \to -\infty$$. Thus, $$\lim_{t \to 6^-} \ln|t-6| - \ln 5 = -\infty$$ 7. Since the first integral diverges to negative infinity, the original integral diverges. 8. Therefore, the integral $$\int_1^\infty \frac{1}{x-6} \, dx$$ diverges and does not have a finite value.