You need to notice that the set A consists of all integer elements `x` that have the property to make the values of the function `f(x)` also integers.

The function `f(x) = (x+3)/(x-1) in Z` if `x - 1` is a divisor of `x + 3` . Since it is not easy to predict what are x values that makes `x - 1` a divisor of `x + 3` , you need to perform the following steps:

`f(x) = (x+3)/(x-1) => f(x) = (x - 1 + 1 + 3)/(x - 1) => f(x) = (x - 1)/(x - 1) + 4/(x - 1)`

`f(x) =1 + 4/(x - 1)`

Considering the condition that `f(x) in Z` , you should notice that the division `4/(x - 1)` `in Z` only if `x - 1` represents one of divisors of `4` , such that:

`x - 1 = 1 => x = 2`

`x - 1 = -2 => x = 0`

`x - 1 = 2 => x = 3`

`x - 1 = -2 => x = -1`

`x - 1 = 4 => x = 5`

`x - 1 = -4 => x = -3`

**Hence, evaluating elements of the set A, under the given conditons, yields `A = {-3;-1;0;2;3;5}` .**

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