There are 2 intervals to consider
I_1=(-oo,-3) and I_2=(-3,+oo)
In the first interval I_1
((-x-3)+x)/(x+2)-1>0
=>, (-3-(x+2))/(x+2)>0
=>, (-x-5)/(x+2)>0
let g(x)=(-x-5)/(x+2)
Build a sign chart
color(white)(aaaa)xcolor(white)(aaaaaa)-oocolor(white)(aaaa)-5color(white)(aaaa)-3
color(white)(aaaa)-x-5color(white)(aaaaa)+color(white)(aaaaa)-
color(white)(aaaa)x+2color(white)(aaaaaaa)-color(white)(aaaaa)-
color(white)(aaaa)g(x)color(white)(aaaaaaaa)-color(white)(aaaaa)+
Therefore, in the interval I_1
g(x)>0 when x in (-5,-3)
In the second interval I_2
((x+3)+x)/(x+2)-1>0
=>, (2x+3-(x+2))/(x+2)>0
=>, (x+1)/(x+2)>0
let h(x)=(x+1)/(x+2)
Build the sign chart
color(white)(aaaa)xcolor(white)(aaaa)-3color(white)(aaaaaa)-2color(white)(aaaaaa)-1color(white)(aaaa)+oo
color(white)(aaaa)x+2color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)x+1color(white)(aaaa)-color(white)(aaaa)#color(white)(aaaaa)-#color(white)(aaaa)+
color(white)(aaaa)h(x)color(white)(aaaaa)+color(white)(aaaa)||color(white)(aaaa)-color(white)(aaaa)+
Therefore, in this interval I_2
h(x)>0 when x in (-3,-2) uu (-1,+oo)
Finally,
The solution is
x in (-5,-3)uu(-3,-2) uu (-1,+oo)
That is
x in (-5,-2)uu(-1,oo)
graph{(|x+3|+x)/(x+2)-1 [-17.21, 11.27, -7.4, 6.84]}