Step-by-Step Solution

1. Let $I = \int\frac{e^{x}}{\sqrt{5-4e^{x}-e^{2x}}}dx$

2. Substitute $u = e^x$, so $du = e^x dx$.

   Then, $I = \int\frac{1}{\sqrt{5-4u-u^{2}}}du = \int\frac{1}{\sqrt{5-(u^{2}+4u)}}du$

3. Complete the square in the denominator: $u^2 + 4u = (u+2)^2 - 4$.

   So, $5 - (u^2 + 4u) = 5 - ((u+2)^2 - 4) = 9 - (u+2)^2$.

   Thus, $I = \int\frac{1}{\sqrt{9-(u+2)^{2}}}du = \int\frac{1}{\sqrt{3^{2}-(u+2)^{2}}}du$

4. Now, use the formula $\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx = \sin^{-1}(\frac{x}{a}) + C$.

   Here, $a = 3$ and $x = u+2$.

   So, $I = \sin^{-1}(\frac{u+2}{3}) + C$

5. Substitute back $u = e^x$.

   Therefore, $I = \sin^{-1}(\frac{e^{x}+2}{3}) + C$