Integral

**jones123** · December 30th 2012, 05:27 PM

Hello! I'm stuck on the following integral: integrate 1 / (1 + ae^x) dx with a = constant term > 0... Thank you!

**skeeter** · December 30th 2012, 05:36 PM

$\frac{1}{1+ae^x} = \frac{1+ae^x -ae^x}{1+ae^x} = \frac{1+ae^x}{1+ae^x} - \frac{ae^x}{1+ae^x} = 1 - \frac{ae^x}{1+ae^x}$

can you integrate the last expression on your own?

**Ahasueros** · December 30th 2012, 05:59 PM

$\int 1/(1+ae^x) dx$ substitute $u = 1+ae^x$ and $du = ae^x dx$
multiply integral by $ae^x/(ae^x)$ , so your integral becomes

$\int 1/(1+ae^x) * ae^x/(ae^x) dx$

using the u substitutions

$\int 1/((u*u-1)) * du$ note: $ae^x = u - 1$

then using partial fractions

$1/((u)(u-1)) = A/u + B/(u-1)<br />$

$1 = A(u-1) + B(u)$

$1 = (A+B)u - A$ -> $A+B = 0$ and $-A= 1$

so $A=-1$ and $B = 1$

your new integral becomes $\int -1/u + 1/(u-1) du$

you should be able to do the rest

Edit: skeeter's way is much faster (didn't see his post before posting this)

**sbhatnagar** · December 30th 2012, 10:17 PM

You can substitute $u=e^x$ .

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

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

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

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

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