Question

Does ln|1-(e^-x)| = -x-ln|(e^s)-1|?

I graphed it but it does not look the same.
The problem is integrate dx/((e^x)-1), I got -x-ln|(e^s-1)| + C as my answer, but the back of the book says it is ln|1-(e^-x)| + C.

Answer 1

Where does the \(s\) come from? If you integrate a function of \(x\) with respect to \(x\), you can only get another function of \(x\).
\[\int\frac{dx}{e^{x}-1}\]
Let \(u=e^{x}-1~\Rightarrow~e^x=u+1\\
du=e^x~dx~\Rightarrow~dx=\dfrac{du}{u+1}\)
\[\int\frac{\frac{du}{u+1}}{u}\\
\int\frac{du}{u(u+1)}\]
Partial fraction decomposition (leaving the details to you):
\[\frac{1}{u(u+1)}=\frac{1}{u}-\frac{1}{u+1}\]
\[\int\frac{du}{u}-\int\frac{du}{u+1}\\
\ln|u|-\ln|u+1|+C\\
\ln|e^x-1|-\ln|e^x|+C\\
\ln|e^x-1|-x+C\]
You seem to be off by a minus sign on the first term.

As for why the book has that answer:
\[\begin{align*}\ln|e^x-1|-\ln|e^x|+C&=\ln\left|\frac{e^x-1}{e^x}\right|+C\\
&=\ln\left|1-e^{-x}\right|+C\end{align*}\]

Answer 2

The s was a typo, lol. I rewrote the numerator as e^s - e^s + 1 and didn't use partial fractions, but I think it was much easier to see it using PF. Thanks!