Step 1: Differentiate the answer
Differentiating the right side, $a\frac{e^{x}}{2\sqrt{e^{x}-1}}+b\frac{1}{1+(e^{x}-1)}\cdot\frac{e^{x}}{2\sqrt{e^{x}-1}}$.
Step 2: Simplify
This equals $\frac{e^{x}}{2\sqrt{e^{x}-1}}\left(a+\frac{b}{e^{x}}\right)=\frac{ae^{x}+b}{2\sqrt{e^{x}-1}}$.
Step 3: Match with the integrand
We need $\sqrt{e^{x}-1}=\frac{e^{x}-1}{\sqrt{e^{x}-1}}$, so $ae^{x}+b=2e^{x}-2$. Hence $a=2,\ b=-2$ and $a+b=0$.
Final Answer:
The correct relation is $a+b=0$.
\[ \boxed{\text{(A) }a+b=0} \]