Hello! I'm stuck on the following integral: integrate 1 / (1 + ae^x) dx with a = constant term > 0... Thank you!
$\displaystyle \int 1/(1+ae^x) dx$ substitute $\displaystyle u = 1+ae^x$ and $\displaystyle du = ae^x dx $
multiply integral by $\displaystyle ae^x/(ae^x)$, so your integral becomes
$\displaystyle \int 1/(1+ae^x) * ae^x/(ae^x) dx$
using the u substitutions
$\displaystyle \int 1/((u*u-1)) * du$ note: $\displaystyle ae^x = u - 1$
then using partial fractions
$\displaystyle 1/((u)(u-1)) = A/u + B/(u-1)
$
$\displaystyle 1 = A(u-1) + B(u)$
$\displaystyle 1 = (A+B)u - A $ -> $\displaystyle A+B = 0 $ and $\displaystyle -A= 1$
so $\displaystyle A=-1$ and$\displaystyle B = 1$
your new integral becomes $\displaystyle \int -1/u + 1/(u-1) du $
you should be able to do the rest
Edit: skeeter's way is much faster (didn't see his post before posting this)
You can substitute $\displaystyle u=e^x$.
$\displaystyle \int \frac{1}{1+ae^x}dx = \int \frac{du}{(1+au)u}$
$\displaystyle =\int \left( \frac{1}{u}-\frac{a}{1+au}\right)du$
$\displaystyle =\ln u -\ln(1+au)+C$
$\displaystyle =x-\ln(1+ae^x)+C$