Integrating Logistic Functions 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?
 A: $$\int \frac{dx}{1+e^{-x}}=\int \frac{e^{x}}{e^{x}}\frac{dx}{1+e^{-x}}$$
$$\int\frac{e^xdx}{e^x+1}=\int\frac{e^x+1-1 dx}{e^x+1}$$
$$\int\frac{e^x+1-1 dx}{e^x+1}=\int\frac{e^x+1 dx}{e^x+1}+\int\frac{-1dx}{e^{x}+1}$$
$$\int\frac{e^x+1 dx}{e^x+1}+\int\frac{-1dx}{e^{x}+1}=\int dx-\int(\frac{e^{-x}}{e^{-x}})\frac{dx}{e^{x}+1}$$
$$\int dx-\int(\frac{e^{-x}}{e^{-x}})\frac{dx}{e^{x}+1}=\int dx -\int\frac{e^{-x}dx}{1+e^{-x}}  $$
$$\int dx -\int\frac{e^{-x}dx}{1+e^{-x}}  =x+\ln|1+e^{-x}|+C$$
Another way to do this would be
$$\int \frac{dx}{1+e^{-x}}=\int \frac{e^{x}}{e^{x}}\frac{dx}{1+e^{-x}}$$
$$\int \frac{e^{x}}{e^{x}}(\frac{dx}{1+e^{-x}})=\int\frac{e^xdx}{e^x+1}$$
$$\int\frac{e^xdx}{e^x+1}=\ln|{e^x+1}|+c$$
A: The trick is to write
$$
\frac{1}{1+e^{-x}}=\frac{1+e^{-x}-e^{-x}}{1+e^{-x}}=\frac{1+e^{-x}}{1+e^{-x}}+\frac{-e^{-x}}{1+e^{-x}}=1+\frac{-e^{-x}}{1+e^{-x}}.
$$
Now you integrate
\begin{align}
\int\frac{1}{1+e^{-x}}dx=\int1+\frac{-e^{-x}}{1+e^{-x}}dx=\int 1dx+\int\frac{-e^{-x}}{1+e^{-x}}dx=x+C+\int\frac{-e^{-x}}{1+e^{-x}}dx
\end{align}
Defining $f(x)=1+e^{-x}$ yields $f'(x)=-e^{-x}$ and
$$
\int\frac{-e^{-x}}{1+e^{-x}}dx=\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C=\ln|1+e^{-x}|+C.
$$
Together we get
$$
\int\frac1{1+e^{-x}}dx=x+\ln|1+e^{-x}|+C
$$
A: 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
$$\int_{x_{1}}^{x_{2}}f(x)dx = F(x_{2}) - F(x_{1})$$
A: 
the denominator is to the power of (−1) and integrating will
  resulting in a power of (0).

This is where you are misunderstood. $\int\frac{1}{x} dx$ = $\ln|x|$. The power rule does not apply when you are integrating $x^{-1}$.
In order to tackle this, you can do u-substitution. You can simplify to:
$$\int \frac{e^x}{e^x+1} dx$$
And you can let $u = e^x+1$, therefore $du = e^x dx$
This leaves you with:
$$\int \frac{1}{u}du$$ And that's the exception to the power rule like I said above.
So this becomes:
$$\ln|u|$$
Substituting back in for $u = e^x+1$ gets you:
$$\ln|e^x+1| + C $$
