Hi,
I am stuck on one integral problem.
How should I find the integral of:
$\displaystyle \int\frac{dx}{1-e^x}\, $
Thanks for your help.
Oh, I see. The negative comes from taking the derivative of e^-x. I was curious; as I was checking my answer using an online integral calculator, they had the answer as It should be $\displaystyle x-ln|e^{-x}-1| + C$. Do you know where they could have gotten the x from? Or is that incorrect?
No.
The answer in the calculator is $\displaystyle x-ln|e^{{\color{red}x}}-1|+C$
Its $\displaystyle e^x$ not $\displaystyle e^{-x}$.
See this:
integrate [ e^[-x] ] / [ e^[-x] - 1 ] - Wolfram|Alpha
Check it please.
It is the same.
If you differentiate this answer and make a common denominator (here you will get a minus sign in the numerator of a fraction) and multiply the resulting fraction by $\displaystyle \frac{e^{-x}}{e^{-x}}$ and take the minus sign as common factor from the denominator and cancel it with the another minus sign, you will get the same function which we integrated it.
I know this maybe will confuse you.
Do the following steps:
1- Differentiate $\displaystyle x-ln|e^x-1|+C$.
2- Make a common denominator to get a fraction with a minus sign in the numerator.
3- Multiply this fraction by $\displaystyle \frac{e^{-x}}{e^{-x}}$.
4- take the minus sign as common factor from the denominator and cancel it with the another one.
5- GOT IT ?