jmc

If $x<-2,$ then $$|x-1|+|x+2|=-(x-1)-(x+2)=-2x-1.$$Solving $-2x-1<5,$ we get $x>-3.$ So, the values of $x$ that work in this case are $-3<x<-2.$

If $-2\le x<1,$ then $$|x-1|+|x+2|=-(x-1)+(x+2)=3.$$All values in $-2\le x<1$ work.

If $1\le x,$ then $$|x-1|+|x+2|=(x-1)+(x+2)=2x+1.$$Solving $2x+1<5,$ we get $x<2.$ So the values of $x$ that work in this case are $1\le x<2.$

Therefore, the solution is $x\in \boxed{(-3,2)}.$