# Integral

• Dec 30th 2012, 05:27 PM
jones123
Integral
Hello! I'm stuck on the following integral: integrate 1 / (1 + ae^x) dx with a = constant term > 0... Thank you!
• Dec 30th 2012, 05:36 PM
skeeter
Re: Integral
$\displaystyle \frac{1}{1+ae^x} = \frac{1+ae^x -ae^x}{1+ae^x} = \frac{1+ae^x}{1+ae^x} - \frac{ae^x}{1+ae^x} = 1 - \frac{ae^x}{1+ae^x}$

can you integrate the last expression on your own?
• Dec 30th 2012, 05:59 PM
Ahasueros
Re: Integral
$\displaystyle \int 1/(1+ae^x) dx$ 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)
• Dec 30th 2012, 10:17 PM
sbhatnagar
Re: Integral
You can substitute $\displaystyle u=e^x$.

$\displaystyle \int \frac{1}{1+ae^x}dx = \int \frac{du}{(1+au)u}$

$\displaystyle =\int \left( \frac{1}{u}-\frac{a}{1+au}\right)du$

$\displaystyle =\ln u -\ln(1+au)+C$

$\displaystyle =x-\ln(1+ae^x)+C$