Solve Integral Error: \int\frac{1+e^x}{1-e^x}dx

[G01]

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

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

The second integral can be done by the substitution
$$u = 1-e^x$$
$$du = -e^x dx$$

So the second integral becomes:
$$\int \frac{du}{u} = \ln|u|+C$$
In the first integral, you can use the substituion:
$$w = e^x$$
$$dw = e^x dx$$

So the first integral becomes:
$$\int \frac{dw}{w(1-w)}$$
This can be done by parts and you get:
$$\int \frac{dw}{w} + \int \frac{dw}{1-w}$$

These are also both natural logs so you end up with:

$$\ln|e^x| - 2\ln|1-e^x| + C$$

$$x - 2\ln|1-e^x| +C$$

I had http://integrals.wolfram.com/index.jsp compute this for me and it got:

$$x - 2\ln(e^x - 1) +C$$

I think this may have something to do with the absolute value, but I always make stupid mistakes with signs so I thought I'd check...Thanks for the time I probably just wasted.

 

[G01]

For some reason it won't let me edit my post... on the fifth line that integral should be
$$-\int\frac{du}{u}$$

 

[benorin]

sure, |a-b|=|b-a|

 

[G01]

ok thanks a lot, stupid question