Integrate (e^x +1)/(e^x -1): Tips & Hints

[Swatch]

Hi. I am trying to integrate (e^x +1)/(e^x -1)
I have looked at this for almost an hour and don´t know how to start with it. I want to use substitution but I have to rewrite this in some way. Could anyone please give me a little hint?

 

[Physics Monkey]

Swatch, try breaking the integral into two pieces and using a different substitution on each piece. One of the pieces will immediately look easy, and the other will be easy after one line of algebra.

 

[HallsofIvy]

Perhaps I'm missing something but a kind of obvious substitution would be
u= e^x. Then (e^x+1)/(e^x-1) becomes (u-1)/(u+ 1). Of course, du= e^xdx so dx= (1/u)du. The integral
$$\int \frac{e^x+1}{e^x-1}dx$$ becomes
$$\int \frac{u+1}{u(u-1)}du$$
which can be done with partial fractions.
If you don't like partial fractions, let u= e^x-1. Then du= e^xdx so dx= (1/u) du again but now e^x+ 1= u+ 2. The integral becomes
$$\int\frac{u+2}{u^2}du= \int \left(u^{-1}+ 2u^{-2}\right)du$$.

 

[Swatch]

I integrated with u=e^x and used partial fraction
to get
-ln(e^x) + 2*ln(e^x -1) +C

I differentiated this back to the beginning, so I should be right. But I got a different looking answer in my textbook.

 

[TD]

If the derivative of your solution gives the integrand again, you should be OK.
It's possible to find different forms of answers when integrating, that very much depends on the method used.

 

[Swatch]

thank you for your help

 

[mathmike]

mathmatica says it is

-2x+xln(x)+Li(x)