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

 

[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.