Although not asked for, it might be insightful for the OP to notice that the logistic equation is in fact a special case of the Bernoulli equation, and can be solved without partial fractions! However it would be advisable to ignore if you (the OP) have not yet covered this :).
The Bernoulli equation is of the form:
$\begin{equation}y' = f(x)y^a - g(x)y \end{equation}$
For convenience, let's jot down the logistic equation also,
$$x' = rx - rx^2$$.
Quite a resemblance, isn't there? Indeed, the Bernoulli equation with $a = 2$ and $f(x) = g(x) = -r$ is the logistic equation, also known as the Riccati equation.
Returning to the logistic equation and dividing through by $x^2$ gives,
$$x' = rx - rx^2 \implies x^{-2} x' = rx^{-1} - r$$.
Now we can make a (perhaps) devious substitution, $u = x^{-1}$.
It follows that, $u' = -x^{-2}x'$.
But we know what $x'$ is, so we can expand this differential equation in $u$,
$u' = -x^{-2}(rx - rx^2) = -r(x^{-1} - 1)$.
Thus, $u' + ru = r ... (1)$
Aha! The differential equation is now in it's linear form. We can use an integrating factor to finish off.
$$I(t) = e^{\int r dt} = e^{rt}.$$
Multiply (1) by $I(t)$,
$e^{rt}u' + e^{rt} ru = e^{rt} r$
We notice the following identity on the left hand side,
$\dfrac{d}{dt}(ue^{rt}) \equiv u'e^{rt} + rue^{rt}$, now we can integrate.$\int \frac{d}{dt}(ue^{rt}) dt = \int re^{rt} dt$
$ue^{rt} = e^{rt} + c$
$u = 1 + ce^{-rt}$
Recall that $u = x^{-1}$,
$x = \dfrac{1}{1 + ce^{-rt}}.$