integrating to clear differential I'm unsure if that's the correct term.
I have an equation $v^2 = \frac{2ILB}{m}x$ and I need to find distance with respect to time (yes, physics.. but the silly math is what trips me up so I'm posting here)
Here's my attempt
$v^2 = [\frac{dx}{dt}]^2 = ...$
$\frac{dx}{dt} = \sqrt{\frac{2ILB}{m}x}$
$\frac{dx}{\sqrt{x}} = dt \sqrt{\frac{2ILB}{m}}$
$2\sqrt{x} = t \sqrt{\frac{2ILB}{m}}$
$x(t) = t^2 \frac{2ILB}{4m}$
My rule of thumb when I don't know if I'm doing something is "well, if I'm not lying at any point how could I be wrong?". I've finished calc3 and honestly never recall dealing with these outside of physics, so I can never know for sure if I'm lying at some point !
Anyways, I'm just wondering if I approached this right and if not, what exactly am I supposed to do?
 A: Your answer is correct.   
Assuming that $v$ is velocity rather than speed, your second line should be  
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$  
If it's speed, your choice to ignore the negative term is ok.  
You should add a constant when you integrate, giving  
$$2\sqrt{x} = \pm t \sqrt{\frac{2ILB}{m}}+C$$  
Also, sometimes (but rarely), squaring both sides of an equation can introduce erroneous solutions. See http://www.jimloy.com/algebra/square.htm for examples. To be certain, you really need to plug solutions back into your original equation to make sure they work, unless you can show analytically that they're ok.  
If $x(0) = 0$, this gives $C=0$.  
Your result is  
$$x(t) = t^2 \frac{2ILB}{4m}$$  
If we check that this satisfies 
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$  
we find the answer is yes in the positive case and no in the negative.  
If $x(0) = 0$, squaring both sides of  
$$2\sqrt{x} = \pm t \sqrt{\frac{2ILB}{m}}+C$$  
gives us your result for $x$.  
We already know that this only satisfies  
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$  
for the positive case and so  
$$\frac{dx}{dt} = \sqrt{\frac{2ILB}{m}x}$$  
As such, your answer  
$$x(t) = t^2 \frac{2ILB}{4m}$$   
is correct.
