help! i forgot how to intergrate exponentials!?

Question

i need to know how to find the integral of (1/e^x)

more specifically ∫1/(e^(x/kt)+1) with respect to x.

thanks a bundle. it's been a while since my calc classes, and integrals have popped up on my p-chem problems.

::EDIT:: well, i guess i'm on the right track, but i ran into an obstacle using subsitution. it doesn't substitute in nicely.

if u = e^(x/kt)+1
then du = e^(x/kt)/(kt^2)dx

yeah...that's where i am right now...

Answer 1

∫1/(e^(x/kt)+1)dx =
let u = x/kt, du = dx/kt
kt∫1/(e^(u)+1)du =
x/kt - ktln(e^(x/kt) + 1)

Answer 2

Integral (1/e^x)dx

To solve this one, all you have to do is realize 1/e^x is equal to e^(-x).

Integral (e^(-x))dx

Using substitution (u = -x), you'll find that this integral is equal to

-e^(-x) + C

As for solving

Integral (1/[e^(x/kt) + 1])dx

I don't quite have the answer so far; sorry.