Intersection of Set theory & Integral Calculus

The above improper integral computes to:

$\Large \frac{1}{\ln 6} \cdot \int_{5}^{\infty} \frac{2}{x-4} - \frac{2}{x+1} dx = \frac{1}{\ln 6} \cdot 2[\ln (x-4) - \ln (x+1)] = \frac{2}{\ln 6} \cdot \ln(\frac{x-4}{x+1}) |_{5}^{\infty} = \frac{2}{\ln 6} \cdot \ln(\frac{1}{1/6}) = \frac{2}{\ln 6} \cdot \ln 6 = 2.$

If $S = 0, \pm 1$ , then the corresponding power set equals:

$(\emptyset); (0); (1); (-1); (-1,0); (0,1); (-1,1); (-1, 0,1)$

which contains $\boxed{8}$ members.