- #1
Sparky_
- 227
- 5
Greetings -
This integral is part of a larger problem I'm working on - I'm stuck (I Think)
Here's the integral:
[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]
I've tried some algebra and substitutions:
[tex] = \int -\frac{e^{2x}}{(1+e^x)}dx[/tex]
[tex] = u=e^x[/tex]
[tex] = du=e^x dx[/tex]
[tex] = \int -\frac{u}{(1+u)}du[/tex]
Is this correct? Is this a standard form for something? - looking like ln()?
OR
could go:
[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]
[tex] = \int -\frac{1}{e^{-2x} +e^{-x}}dx[/tex]
This integral is part of a larger problem I'm working on - I'm stuck (I Think)
Here's the integral:
[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]
I've tried some algebra and substitutions:
[tex] = \int -\frac{e^{2x}}{(1+e^x)}dx[/tex]
[tex] = u=e^x[/tex]
[tex] = du=e^x dx[/tex]
[tex] = \int -\frac{u}{(1+u)}du[/tex]
Is this correct? Is this a standard form for something? - looking like ln()?
OR
could go:
[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]
[tex] = \int -\frac{1}{e^{-2x} +e^{-x}}dx[/tex]
Last edited: