Integration

Integration

Postby markosheehan » Sat Sep 03, 2016 8:54 am

how do you integrate e^X-e^-x/e^-x+1 dx
markosheehan
 
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Re: integration

Postby Guest » Sun Sep 04, 2016 4:57 am

[tex]\frac{e^x-e^{-x}}{e^{-x}+1}[/tex]
[tex]= -1+\frac{e^x+1}{e^{-x}+1}[/tex]
[tex]=-1+\frac{e^x(1+e^{-x})}{e^{-x}+1}[/tex]
[tex]=-1+e^x[/tex]
which should be easy to integrate.

Hope this helped,

R. Baber.
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Re: integration

Postby markosheehan » Sun Sep 04, 2016 5:23 am

how did you do this did you multiply by something or did you let -e^-1=1.

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Re: integration

Postby Guest » Sun Sep 04, 2016 6:35 am

There are many ways you could use to simplify the expression, one such way is to use the substitution [tex]u=e^x[/tex]
So
[tex]\frac{e^x-e^{-x}}{e^{-x}+1}[/tex]
[tex]=\frac{u-\frac{1}{u}}{\frac{1}{u}+1}[/tex]
[tex]=\frac{(u-\frac{1}{u})\times u}{(\frac{1}{u}+1)\times u}[/tex]
[tex]=\frac{u^2-1}{1+u}[/tex]
[tex]=\frac{(u-1)(u+1)}{1+u}[/tex]
[tex]=u-1[/tex]
[tex]=e^x-1[/tex]

Hope this helped,

R. Baber.
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