What is the Integral with e^x in it?

[Outlaw747]

Homework Statement

\int\frac{1+e^{x}}{1-e^{x}} dx

Homework Equations

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The Attempt at a Solution

I've tried substituting for u=e^x or u=1-e^x but I can't seem to get anywhere. Haven't done calc in a while and just want someone to point me in the right direction. Thanks.

 

[loveequation]

Letting u = e^x [/tex] gives<br /> &lt;br /&gt; \int\frac{1+u}{1-u}\frac{1}{u}\,du&lt;br /&gt;<br /> Next expand in partial fractions. The integrand becomes<br /> &lt;br /&gt; \frac{2}{1-u} + \frac{1}{u}&lt;br /&gt;<br /> which you can easily integrate.

 

[PhaseShifter]

I'm thinking u=e^{x} might work, at least for x<0.

Also, {{1+u}\over{1-u}}={{1-u+2u}\over{1-u}}=1+2{{u}\over{1-u}}

 

[PhaseShifter]

For x>0, you might try u=e^{-x}

{{1+e^{x}}\over{1-e^{x}}}={{e^{-x}+1}\over{e^{-x}-1}}={{u+1}\over{u-1}}={{2u}\over{u-1}}-{{u-1}\over{u-1}}
\int({1\over{u}}-{2\over{u-1}})du

 

[VietDao29]

For x>0, you might try u=e^{-x}

{{1+e^{x}}\over{1-e^{x}}}={{e^{-x}+1}\over{e^{-x}-1}}={{u+1}\over{u-1}}={{2u}\over{u-1}}-{{u-1}\over{u-1}}



\int({1\over{u}}-{2\over{u-1}})du

Well, why can't you sub u = e^x, when x > 0?

 

[HallsofIvy]

You can- it really doesn't make any difference.

 

[Outlaw747]

Thanks a lot guys.