TITLE: Integrating Logistic Functions
QUESTION [9 upvotes]: For logistic functions in the form of $\frac{C}{1+Ae^{-bx}}$ where $C,A,b>0$ and $x$ is the independent variable, how does one integrate this function type? since during integration, the denominator is to the power of $(-1)$ and integrating will resulting in a power of $(0)$. I have tried a few websites such as cymath, wolfram and symbolab but havent been able to understand their working. for example 
$$\int \frac{1}{1+e^{-x}}dx=x+\ln \left|1+e^{-x}\right|+C$$. 
edit: if the anti-derivative is used for finding an area of the original function through integration, how can an unknown bound for a particular area be solved? is it possible?

REPLY [2 votes]: If we have $f(x) = \frac{C}{1 + Ae^{-bx}}$, with $C,A,b > 0$. We can find its antiderivative
$$F(x) = \int \frac{C}{1 + Ae^{-bx}}dx$$
$$= C\int \frac{1}{1 + Ae^{-bx}}dx$$
$$= C\int \frac{1 + Ae^{-bx} - Ae^{-bx}}{1 + Ae^{-bx}}dx$$
$$= C\int 1 - \frac{Ae^{-bx}}{1 + Ae^{-bx}}dx$$
$$= C\int dx - C\int \frac{Ae^{-bx}}{1 + Ae^{-bx}}dx$$
$$= Cx - C\frac{1}{-b} \int \frac{-bAe^{-bx}}{1 + Ae^{-bx}}dx$$
$$= Cx + \frac{C}{b} \int \frac{(1 + Ae^{-bx})'}{1 + Ae^{-bx}}dx$$
$$= Cx + \frac{C}{b} ln|1 + Ae^{-bx}| + Const$$
Now, if you want to know what is the area under the curve, you just need compute
$$\int_{x_{1}}^{x_{2}}f(x)dx = F(x_{2}) - F(x_{1})$$