how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1))

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cathy_liping
Posts: 9
Joined: Sat Oct 01, 2011 11:21 pm

how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1))

Post by cathy_liping » Sat Oct 08, 2011 5:50 am

how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1)) where x is from 0 to 1. i have no idea when i saw it... tks in advance!

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1))

Post by Hom » Sat Oct 08, 2011 9:15 am

cathy_liping wrote:how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1)) where x is from 0 to 1. i have no idea when i saw it... tks in advance!
Try to use the LaTex format next time. I found this site quite helpful to edit the formulae in real time. http://www.codecogs.com/latex/eqneditor.php

Well. $$\int_{0}^{1} \frac{(e^{ax}-e^{bx})} {(e^{ax}+1)(e^{bx}+1)} dx = \int_{0}^{1} (\frac{e^{ax}} {(e^{ax}+1)} - \frac{e^{bx}} {(e^{bx}+1)}) dx$$
from here you can substitution $$u = e^{ax}$$ so $$du= ae^{ax} dx$$. After some manipulation, you will get $$\frac{ log(e^{ax}+1)}{a}-\frac{ (e^{bx}+1)}{b}$$.

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DDswife
Posts: 161
Joined: Thu Aug 14, 2014 5:29 pm

Re: how to integrate (exp(ax)-exp(bx))/((exp(ax)+1)(exp(bx)+1))

Post by DDswife » Sat Aug 23, 2014 3:54 pm

I think that you forgot to write log in the second term



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