How Do You Integrate $\int \frac{1}{1-e^x} dx$ with Substitution?

[karush]

$\int \frac{1}{1-e^x} dx$
$u=1-e^x,\ du=-e^x dx$
Not sure of next step

 

[topsquark]

$\int \frac{1}{1-e^x} dx$
$u=1-e^x,\ du=-e^x dx$
Not sure of next step

Try this:
[math]\int \frac{1}{1 - e^x}~dx = \int \frac{1 - e^x + e^x}{1 - e^x}~dx[/math]

[math]= \int \frac{1 - e^x}{1 - e^x}~dx + \int \frac{e^x}{1 - e^x}~dx[/math]

-Dan

 

[evinda]

$\int \frac{1}{1-e^x} dx$
$u=1-e^x,\ du=-e^x dx$
Not sure of next step

$$u=1-e^x \\ du=-e^x dx \Rightarrow dx=-\frac{du}{e^x}=\frac{du}{u-1}$$
$$\int \frac{1}{1-e^x}dx=\int \frac{1}{u} \frac{1}{u-1}du=\int \frac{1}{u(u-1)}du=(*)$$

$$\frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{u-1}=\frac{A(u-1)+Bu}{u(u-1)}=\frac{(A+B)u-A}{u(u-1)}$$

So,it must be:

$$A+B=0 \Rightarrow B=-A \\ -A=1 \Rightarrow A=-1 \\ B=-A=1$$

$$(*)=\int \left ( -\frac{1}{u}+\frac{1}{u-1}\right )du=- \ln | u |+\ln{|u-1|}+c \\ =- \ln |1-e^x|+\ln |1-e^x-1|+c=-\ln |1-e^x|+x+c$$

 

[karush]

Cool trick ... Thanks 4 help...I got
$$x-ln|e^x-1|+C$$

 

[evinda]

Cool trick ... Thanks 4 help...I got
$$x-ln|e^x-1|+C$$

Nice! So, we got the same result, since $\ln |1-e^x|=\ln |e^x-1|$.. (Smirk)

 

[mathbalarka]

There are worse methods of doing it :

$$\begin{aligned}\int \frac{1}{1-e^x}, \mathrm{d}x = \int \left ( 1 + e^x + e^{2x} + e^{3x} + \cdots \right ) \, \mathrm{d}x &= \left ( x + e^x + \frac{e^{2x}}{2} + \frac{e^{3x}}{3} + \cdots \right ) + C \\ &= x + \sum_{n = 1}^\infty \frac{(e^x)^n}{n} + C \\ &= x - \log(1 - e^x) + C\end{aligned}$$

So, yes, karush, your answer agrees with mine.

 

[MarkFL]

This is what I would do:

$$\int\frac{1}{1-e^x}\,dx=-\int\frac{-e^{-x}}{e^{-x}-1}\,dx=-\ln\left|e^{-x}-1\right|+C$$

 

[karush]

This is what I would do:

$$\int\frac{1}{1-e^x}\,dx=-\int\frac{-e^{-x}}{e^{-x}-1}\,dx=-\ln\left|e^{-x}-1\right|+C$$

Isn't there another x in the answer?

 

[MarkFL]

Isn't there another x in the answer?

Try rewriting the log argument in terms of $e^x$ and see what happens...:D

 

[karush]

OK I see the x appears...