How to solve this IVP? Could you please help me solve this IVP? 
A certain population grows according to the diﬀerential equation: 
$$\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{P}{20}\left(1 − \frac{P}{4000}\right) $$
and the initial condition $P(0) = 1000$. What is the size of the population at time $t = 10$? 
The answer is
$$P(10)=\frac{4000}{(1 + 3e^{1/2})}$$
but I can't seem to get it. I have tried to bring all the right hand side terms to integrate with respect to $P$. But I end up getting stuck with this equation $$20\ln{(P)} + \frac{80000}{P} = 90 + 20\ln{(1000)}$$ and don't know how to isolate $P$...
Thanks so much! :)

$\textbf{Given}$
$\dfrac{dP}{dt} = \dfrac{P}{20}\left(1-\dfrac{P}{4000}\right)$, $P(0) = 1000$. 
$\textbf{Find}$ 
$P(10)$
$\textbf{Analysis}$
$\dfrac{dP}{dt} = \dfrac{P}{20}\left(1-\dfrac{P}{4000}\right)$.
First divide out $P\left(1-\dfrac{P}{4000}\right)$ from the RHS and multiply $dt$ over from the LHS: 
$$
\dfrac{dP}{P(1-P/4000)} = \dfrac{dt}{20}.  \hspace{3in} (1)
$$
Next use partial fraction decomposition on the LHS: 
\begin{align*}
\dfrac{dP}{P(1-P/4000)} & = \dfrac{A\ dP}{P} + \dfrac{B\ dP}{(1-P/4000)} \\
                        & \Rightarrow 1 = A\left(1-\dfrac{P}{4000}\right) + BP\\ 
                        & \Rightarrow \left\{\begin{array}{lr}
                                              A = 1\\
                                              B =  1/4000
                                      \end{array}\right.
\end{align*}
The solutions for $A,B$ comes from equating coefficients. 
Hence, we may rewrite equation (1) as 
$$
\dfrac{dP}{P} + \dfrac{dP}{4000-P} = \dfrac{dt}{20}.
$$ 
Integrate both sides 
$$
\int \left(\dfrac{1}{P} + \dfrac{1}{4000-P}\right)dP = \dfrac{1}{20}\int dt,
$$
$$
\ln\left(\dfrac{P}{4000-P}\right) = \dfrac{t}{20} + C.
$$
Exponentiate both sides 
$$
\dfrac{P}{4000-P} = Ke^{t/20} \hspace{1cm} \Rightarrow  \hspace{1cm} P = (4000-P)Ke^{t/20}
$$
so that
$$
P = 4000\dfrac{Ke^{t/20}}{1 + Ke^{t/20}}.
$$
Now, the initial condition is $P(0) = 1000$, thus 
$$
P(0) \equiv 1000 = 4000\dfrac{K}{1 + K} \hspace{1cm} \Rightarrow \hspace{1in} K=\dfrac{1}{3}.
$$
So $P(t)$ is given explicitly by 
$$
P(t) = \dfrac{4000}{1+3e^{t/20}},
$$
and $P(10) = \dfrac{4000}{1+3e^{-1/2}}$.
