Trying to understand an integral algebraically and conceptually $f'(t) = \frac{Ae^t}{(0.02A+e^t)^2}$ It is defined that $f'(t) = \frac{Ae^t}{(0.02A+e^t)^2}$ is the rate of population.
It is also given that the population at $t=0$ is given as 6.
Our goal is to find the time $t$ where the population hits $30$.
I want to solve of the constant $A$, but I am not sure how to.
My understanding is that in order to do so, we just take the integral of $f'(t)$ and plug in $f(0)=6$, but I ended up only to find the value of the constant that results from taking any indefinite integral.
Can someone help me out ?
According to the book I have $A = 46.15$ by finding $\lim_{t \to \infty}f(t)=30$.
Which I couldn't understand conceptually why we can do so.

Well,
$$f(t)-f(0) = A \int_0^t dt' \frac{e^{t'}}{(0.02 A+e^{t'})^2}$$
To evaluate the integral, substitute $u=e^{t'}$ and get
$$f(t)-6 = A \int_1^{e^{t}} \frac{du}{(u+0.02 A)^2}  = A \left [\frac{1}{1+0.02 A}-\frac{1}{e^t+0.02 A} \right ]$$
Now you gave us
$$\lim_{t \to \infty} f(t)=30$$
so that
$$30-6=24=\frac{A}{1+0.02 A} \implies 24 + \frac{12}{25} A=A \implies A = \frac{600}{13} \approx 46.15$$
