Simple Integration

1. Oct 25, 2005

Swatch

Hi. Im 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?

2. Oct 25, 2005

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.

3. Oct 25, 2005

HallsofIvy

Staff Emeritus
Perhaps I'm missing something but a kind of obvious substitution would be
u= ex. Then (ex+1)/(ex-1) becomes (u-1)/(u+ 1). Of course, du= exdx 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= ex-1. Then du= exdx so dx= (1/u) du again but now ex+ 1= u+ 2. The integral becomes
$$\int\frac{u+2}{u^2}du= \int \left(u^{-1}+ 2u^{-2}\right)du$$.

Last edited: Oct 25, 2005
4. Oct 26, 2005

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.

5. Oct 26, 2005

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.

6. Oct 26, 2005

Swatch

7. Oct 26, 2005

mathmike

mathmatica says it is

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