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Correct answer: (-infinity, -3) U (-1, 1] U [2, infinity)
We need (x-1)(x-2)/((x+1)(x+3)) >= 0 with (x+1)(x+3) != 0. Critical points: x = -3, -1 (excluded, denominator zero) and x = 1, 2 (allowed since numerator can be zero). Sign analysis of the rational function across intervals gives the expression >= 0 on (-inf,-3), on (-1,1], and on [2,inf). At x = -3 and x = -1 the denominator is zero (excluded). At x = 1 and x = 2 the value is 0 (allowed).