# integration involving exponentials

• Apr 26th 2010, 10:11 AM
wahhdoe
integration involving exponentials
Hi, I have a question and the answer but i am slightly uncertain how the answer is achieved and was wondering if someone could please explain it to me.

Q: Using the fact that 2/(e^[2x]+1)=1 - (e^[2x]-1)/(e^[2x]+1)

integrate 1/(e^[2x]+1) with respect to x

A: 1/2{x-ln(e^x+e^-x)}+c

any help would be great thanks
• Apr 26th 2010, 10:33 AM
Failure
Quote:

Originally Posted by wahhdoe
Hi, I have a question and the answer but i am slightly uncertain how the answer is achieved and was wondering if someone could please explain it to me.

Q: Using the fact that 2/(e^[2x]+1)=1 - (e^[2x]-1)/(e^[2x]+1)

integrate 1/(e^[2x]+1) with respect to x

A: 1/2{x-ln(e^x+e^-x)}+c

any help would be great thanks

I don't think that any particularly clever idea is required to solve this integral, just substitute $\displaystyle z := e^{2x}+1$, thus $\displaystyle dz=2e^{2x}\,dx$ and therefore $\displaystyle dx=\frac{dz}{2(z-1)}$.
This gives

$\displaystyle \int \frac{1}{e^{2x}+1}\, dx=\frac{1}{2}\int \frac{dz}{z(z-1)}=\frac{1}{2}\int\left[\frac{1}{z-1}-\frac{1}{z}\right]\,dz=\ldots$
• Apr 26th 2010, 10:39 AM
tom@ballooncalculus
Although perhaps they meant...

$\displaystyle \frac{1}{e^{2x} + 1}$

$\displaystyle =\ \frac{1}{2}\ [1\ -\ \frac{e^{2x} - 1}{e^{2x} + 1}]$

$\displaystyle =\ \frac{1}{2}\ [1\ -\ \frac{e^x - e^{-x}}{e^{x} + e^{-x}}]$

And now the top is the derivative of the bottom. So you can do...

http://www.ballooncalculus.org/asy/intChain/exp.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift being...

http://www.ballooncalculus.org/asy/maps/intChain.png
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