Answer
$\ln |1-e^{-s}|+c$
Work Step by Step
Consider the integral $\int \dfrac{ds}{e^s-1}$
Let us take the help of the substitution method.
$e^s-1=a \implies da=e^s ds; ds=\dfrac{da}{x+1}$
The given integral can be re-written as: $\int \dfrac{ds}{e^s-1}=\int \dfrac{da}{a(a+1)}$
This implies that $\int \dfrac{[(a+1)-a]da}{a(a+1)}=\int [\dfrac{1}{a}- \dfrac{1}{a+1} ]da=\ln | \dfrac{a}{1+a}|+C$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
or, $\ln |\dfrac{e^s-1}{e^s-1+1}|+c=\ln |1-e^{-s}|+c$
