To determine if the integral is convergent or divergent, we need to evaluate the integral up to the vertical asymptote of the integrand and then take the limit to the asymptote.

`int_2^8 1/(x-6)^3dx` let `u=x-6` then `du=dx` and the limits become -4 and 2

`=int_{-4}^2 u^{-3}du`

`=lim_{epsilon->0}(int_{-4}^epsilon u^{-3}du)+lim_{epsilon->0}(int_epsilon^2 u^{-3}du)`

`=lim_{epsilon->0}(-1/2u^{-2}|_{-4}^epsilon)+lim_{epsilon->0}(-1/2 u^{-2}|_epsilon^2)`

`=-1/2lim_{epsilon->0}(1/epsilon^2-1/16)-1/2lim_{epsilon->0}(1/4-1/epsilon^2)`

**Since each part is divergent to negative infinity, the integral is divergent to negative infinity (MINF).**