Integrating with the substitution $u = N^{-v}$ 
Integrate
$$\frac{dN}{dt} = rN(1 - K^{-v}N^v)$$
where $r,v,K$ are positive constants (and so I think $N$ is a function of $t$), using the substitution $u =N^{-v}$, given that $N$ has an initial value at $N_0 < K$. Determine the behaviour of the solution for large times.

So, in accordance with the answers, I have done everything correct and got to the point where I now have
$$\frac{du}{dt} = -vr(u - K^{-v}),$$
but now I'm a little stuck. For some reason, they have integrated
$$\int_{u_0}^u \frac{du}{u - K^{-v}} = -vrt.$$
Why have they integrated between those two limits as opposed to ingetrating just $\frac{du}{u - K^{-v}}$ and then solving for $u$ to get the initial condition like normal?
 A: Because integrating with the lower limit $u=u_0$ is equivalent to applying the initial condition at $t=0$.  To see this, write instead
$$\int_{u_0}^u \frac{du'}{u'-K^{-v}} = -v r \int_0^t dt'$$
Note that I primed the integration variables; they are dummy variables and what we call them is of no importance to the out come of the problem.  The limits, however, are important: the lower limits and upper limits of integration on each side of the equation must correspond.  That is, $u(0)=u_0$ which is equivalent to $N(0)=N_0$.
You can also see this by integrating as you would expect, an indefinite integral:
$$\int \frac{du}{u-K^{-v}} = -v r t + C$$
where $C$ is an integration constant.  This of course implies that
$$\log{(u-K^{-v})} = -v r t + C$$
Now, at $t=0$, $u=u_0$, which means that
$$C = \log{(u_0-K^{-v})}$$
so that
$$\log{(u-K^{-v})}-\log{(u_0-K^{-v})} = -v r t$$
which is equivalent to
$$\int_{u_0}^u \frac{du'}{u'-K^{-v}} = -v r t$$
I hope this makes sense of where the lower integration limit comes from.
