Answer
$\ln |\dfrac{\sqrt {e^s+1}-1}{\sqrt {e^s+1}+1}|+c$
Work Step by Step
Consider the integral $\int \dfrac{ds}{\sqrt {e^s+1}}$
Let us take the help of the substitution method.
$\sqrt {e^s+1}=a \implies da=e^s ds; ds=\dfrac{(2a) da}{a^2-1}$
The given integral can be re-written as: $\int \dfrac{2da}{a^2-1}=\int \dfrac{(a+1)-(a-1)}{(a-1)(a+1)}da$
This implies that $\int \dfrac{1}{a-1} da- \int \dfrac{1}{a+1} ]da=\ln | \dfrac{a-1}{a+1}|+C$
or, $=\ln |\dfrac{\sqrt {e^s+1}-1}{\sqrt {e^s+1}+1}|+c$
