An elementary calculus text defines the "logistic function" as the following:
$$\int \frac{1}{P(1 - mP)} \, dP$$
The solution given is $$\ln|\frac{P}{(1 - mP)}| $$
My incorrect solution is the following:
$$ \frac{1}{P(1 - mP)} = \frac{A}{P} + \frac{B}{1 -mP} $$
=> $$1 = A - Amp + BP $$ => $$ A = 1 $$ and $$ -(1)mP = BP = 0$$ => $$p(-m + B) = 0$$ => $$B = m$$
Therefore,
$$\int \frac{1}{P(1 - mP)} \, dP = \int \frac{1}{P} dP + \int \frac{m}{1 - mp} dP = ln|P| + m(ln|1 -mP|) + C$$
...which is not quite right.
It's possible my eyes are glazing over something obvious. Either way, please let me know which step is fallacious.
Thanks.
The problem is very simple - your integration $\int \frac{m}{1 - mP}dP$. You need $u$ substitution. If $u = 1 - mP$, then $du = -m\,dP$. However, the numerator is just $m\,dP$ so we need to move the negative over: $-du = m\,dP$. Then this becomes:
$$-\int\frac{du}{u} = -\ln|u| = -\ln|1 - mP|$$
Incorporating this into your final answer gives:
$$\ln|P| - \ln|1 - mP| + C$$
Logarithm rules then apply to transform to the final result (you can take subtracted logarithms and convert them to divisors inside the logarithm function):
$$\ln\,\biggr|\frac{P}{1 - mP}\biggr| + C$$