Could someone give me a leg-up for this question?
Is the range restricted to x >= -1? (In order to have only real solutions)?
If so then you should consider the intervals [-1,0) and [0,whatever]. Also you will have to put a constraint on the limits if you want a finite answer.
Recall that x(x+1)^(1/2) is an increasing function which should make your job easier.
Basically the signum function is -1 if x < 0, 0 if x = 0 and 1 if x > 0.
You need to find when the function is < 0, 0, and > 0.
x(x+1)^(1/2) = 0 implies x = 0 or x = -1.
This means sig (x(x+1)^1/2) when x > 0 = 1 and this means Integral 0 to u sig(x(x+1)^(1/2)) = Integral 0 to u 1dx = u - 0 = u.
Now you need to find when x(x+1)^(1/2) < 0 and for that region use sig(x(x+1)^(1/2)) = -1 when the condition is satisfied.
Forget that way: just find when the expression inside the signum function is positive, negative, and 0 and replace it with +1, -1, and 0 respectively and integrate it piece-wise. (For example if u is positive for sig(u) then replace sig(u(x))dx with +1dx.)