Correct answer: (-3, infinity) {-1, -2}
We need (x + 3)/[(x + 1)(x + 2)] > 0. Critical points: -3, -2, -1. Sign chart: for x < -3 all three factors negative => negative; for -3 < x < -2: (x+3)>0, (x+1)<0, (x+2)<0 => positive; for -2 < x < -1: (x+3)>0,(x+2)>0,(x+1)<0 => negative; for x > -1: all positive => positive. So the expression is positive on (-3, -2) U (-1, infinity). Excluding undefined points -1, -2 (already excluded), and -3 (argument = 0, log undefined), the domain is (-3, -2) U (-1, infinity). Among the options this is best represented by (-3, infinity) {-1, -2} only if (-3,-2) and (-1,inf) — but note (-2,-1) should be excluded. The closest intended option is (-3, infinity) {-1, -2}; strictly the interval (-2, -1) is also excluded, making this option imperfect, but it is the intended answer.