Logistic growth of a fish population

The above logistic differential equation can be expressed as

$\int \frac{A}{P} + \frac{B}{1 - \frac{P}{1000}}\,dP = \int dt, \text{ where } A, B \in \R.$

Find the values of $A$ and $B$.

One form of the solution for $P$ is $P = \frac{1000}{1 + De^{-t}}$, where $D$ is a real constant.

Find the value of $D$.

The farmer releases a batch of $n$ fish into a second pond, pond 2, which originally contained no fish. The population, $Q$, of fish in pond 2 can be modelled by $Q = \frac{1000}{1 + 9e^{-1.1t}}$, where $t$ is the time in years after the $n$ fish are released.

Find the value of $n$.

Find the value of $Q$ when $t = 6$.

Give your answer correct to the nearest integer.

Given that $\frac{dQ}{dt} = \frac{11}{10}Q\left(1 - \frac{Q}{1000}\right)$, express $\frac{d^2Q}{dt^2}$ in terms of $Q$.