Say you are working with acceleration as a function of displacement and you are using calculus. $a = \frac{1}{(s - 600)^2}$. If you wanted to obtain velocity you'd use $a = v\frac{dv}{ds}$ so $vdv = \frac{1}{(s-600)^2} ds$.
I can integrate the left-hand side no problem, but I am having trouble with the right-hand side. My first instinct was to try a u-substitution $u = s - 600$, but I ended up with $\frac{-2}{(s-600)^3}$ when the correct answer is $-\frac{1}{s-600}$.
How do I integrate it?
You had the right idea, just did the integral wrong. Using $u=s-600$, we see $du=ds$ so $\int{\frac{1}{(s-600)^2}ds}=\int{\frac{1}{u^2}du}=\int{u^{-2}du}=-u^{-1}=\frac{-1}{s-600}$