# Integral with e^x in it

1. Sep 6, 2009

### Outlaw747

1. The problem statement, all variables and given/known data
$$\int\frac{1+e^{x}}{1-e^{x}}$$ dx

2. Relevant equations
.

3. 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.

Last edited: Sep 6, 2009
2. Sep 6, 2009

### loveequation

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

Last edited: Sep 7, 2009
3. Sep 6, 2009

### PhaseShifter

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

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

Last edited: Sep 6, 2009
4. Sep 6, 2009

### PhaseShifter

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

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

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

Last edited: Sep 6, 2009
5. Sep 7, 2009

### VietDao29

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

6. Sep 7, 2009

### HallsofIvy

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

7. Sep 7, 2009

### Outlaw747

Thanks a lot guys.