Help on population differential equation 
$$\frac{\mathrm dr}{\mathrm dt} = kr \left(1-\frac{r}{r_*}\right) -\alpha fr,$$
  where $k>0$ is a constant representing the rabbit breeding rate, $r_*>0$ is the (constant) maximun sustainable rabbit population size in the absence of predation, $f>0$ is the population of foxes, and $\alpha>0$ is the (constant) rate or predation of rabbits by foxes.

I am told that the fox population is constant.
I'm trying to solve this differential equation. I tried using the Bernoulli equation method, however it did not work out, which is shown here (habbit of mine is I accidently made $r_*$ that was for the maximum sustainable rabbit population size in the absence of predation later into $A$ since it was easier:)
$$\frac{\mathrm dr}{\mathrm dt} = kr -\frac{kr^2}{r_*} -\alpha fr$$
Define $w(t)$ by $w=r^{1-2}=\frac{1}{r}$, so that $$\frac{\mathrm dw}{\mathrm dt} = -\frac{1}{r^2} \frac{\mathrm dr}{\mathrm dt}$$
$$\implies \frac{\mathrm dw}{\mathrm dt} +kw = \frac{k}{r_*} +\alpha fw,$$
where this would become $$e^{kt} w(t) = \frac{e^{kt}}{A} +\frac{\alpha fwe^{kt}}{k} +c.$$
As I made $m(t)=k$, leading to $e^{\int k} = e^{kt}$ and the left hand side was inverse product rule and the right hand side was integration
$$ \implies w(t) =\frac{1}{A} +\frac{\alpha fw}{k} +ce^{-kt}$$
$$\implies r(t) = A +\frac{k}{\alpha fw} +\frac{e^{kt}}{c}$$
 A: The equation is separable.  To better see this, start by multiplying everything out and collecting up the constants:
$$\frac{\mathrm dr}{\mathrm dt}
= kr \left(1-\frac{r}{r_*}\right) -\alpha fr
= kr - \frac{kr^2}{r_*} - \alpha f r
= \underbrace{-\frac{k}{r_*}}_{A}r^2 + \underbrace{(k-\alpha f)}_{B}r
= Ar^2 + Br,$$
which is more clearly separable.  In particular (by a standard abuse of notation), it may be rewritten as
$$
\frac{1}{Ar^2 + Br} \,\mathrm{d}r = \mathrm{d}t
\implies \int \frac{1}{Ar^2 + Br} \,\mathrm{d}r = \int \mathrm{d}t.
$$
As far as the differential equation is concerned, I would say that we are basically done at this point.  However, we probably want something a little more explicit, so maybe we ought to carry out the integration.
The right-hand side is relatively easy to work with, so let's ignore that for now.  On the left-hand side, complete the square in the denominator to get
$$
\int \frac{1}{Ar^2 + Br} \,\mathrm{d}r
= \int \frac{1}{A\left(r + \frac{B}{2A}\right)^2 - \frac{B^2}{4A}}\,\mathrm{d}r.
$$
Make the change of variables $s = r + B/2A$ to obtaina
$$
\int \frac{1}{A\left(r + \frac{B}{2A}\right)^2 - \frac{B^2}{4A}}\,\mathrm{d}r
= \int \frac{1}{As^2 - \frac{B^2}{4A}}\,\mathrm{d}s
= \frac{1}{A} \int \frac{1}{s^2 - \left(\frac{B}{2A}\right)^2}\,\mathrm{d}s
= \frac{1}{A} \int \frac{1}{s^2 - C^2}\,\mathrm{d}s,
$$
where $C = B/2A$ (gathering up the constants so that we don't have to deal with them).  While there are more clever approaches that one might take here, I think that cranking out the partial fractions is the most elementary thing which we could do.  This gives us
\begin{align}
\frac{1}{A} \int \frac{1}{s^2 - C^2}\,\mathrm{d}s
&= \frac{1}{A} \int \frac{1}{2C(s-C)} - \frac{1}{2C(s+C)} \,\mathrm{d}s \\
&= \frac{1}{2AC} \int \frac{1}{s-C} - \frac{1}{s+C} \,\mathrm{d}s \\
&= \frac{1}{2AC}\left( \log(s-C) - \log(s+C) \right) \\
&= \frac{1}{2AC} \log\left( \frac{s-C}{s+C} \right),
\end{align}
where I am assuming (in the last step) that $s > |C|$.  This is a reasonable assumption, as $s$ is related to the population of rabbits, which itself must be a positive number.  At this point, the original right-hand integral has been evaluated.  The next step is to restore the original variable $r$, using the fact that $s = r+C$.
\begin{align}
t + K
&= \int \,\mathrm{d}t \tag{$K$ is the constant of integration} \\
&= \int \frac{1}{Ar^2 + Br} \,\mathrm{d}r \\
&= \frac{1}{2AC} \int \frac{1}{s-C} - \frac{1}{s+C} \,\mathrm{d}S \\
&= \frac{1}{2AC} \log\left( \frac{s-C}{s+C} \right) \\
&= \frac{1}{2AC} \log\left( \frac{r}{r+\frac{B}{A}} \right).
\end{align}
Since the goal is to determine the rabbit population $r$, we solve to get
\begin{align}
&t + K = \frac{1}{2AC} \log\left( \frac{r}{r+\frac{B}{A}} \right) \\
&\qquad\implies 2ACt + K = \log\left( \frac{r}{r+\frac{B}{A}} \right) \tag{note: this is not the same $K$} \\
&\qquad\implies K \mathrm{e}^{2ACt} = \frac{r}{r+\frac{B}{A}} \tag{$K$ changed again} \\
&\qquad\implies \left( K \mathrm{e}^{2ACt} - 1 \right)r = -\frac{B}{A}K \mathrm{e}^{2ACt} \\
&\qquad\implies r = \frac{B}{A}K \left( \frac{\mathrm{e}^{2ACt}}{1 - K \mathrm{e}^{2ACt}} \right) = -\frac{B}{A} \left( \frac{\mathrm{e}^{2ACt}}{K + \mathrm{e}^{2ACt}} \right). \tag{$K$ changes again here}
\end{align}
Finally, restoring the original constants from
$$ A = -\frac{k}{r_*},
\quad
B = k-\alpha f,
\quad\text{and}\quad
2AC = 2A\frac{B}{2A} = B = k-\alpha f, $$
we obtain the solution
$$
\boxed{r(t) = \frac{r_*(k-\alpha f)}{k} \left( \frac{\mathrm{e}^{(k-\alpha f)t}}{K + \mathrm{e}^{(k-\alpha f)t}} \right).}
$$
I am sure that there are some silly arithmetic errors in here somewhere, but I'll be damned if I can find them.  In any event, I built a Desmos demonstration which can be used to play around with this a bit.  The curves that I get for "reasonable" values of the constants look like logistic growth curves, which is what we would expect.  So, I think I did something right.
