Answer
$\displaystyle \frac{2}{3}\sqrt{3e^{x}-1}+C$
Work Step by Step
$\displaystyle \int\frac{e^{x}}{\sqrt{3e^{x}-1}}dx=\int e^{x}(3e^{x}-1)^{-1/2}$
Shortcut to apply:
$\displaystyle \quad\int g\cdot u^{n}dx=\frac{g}{u'}\cdot\frac{u^{n+1}}{n+1}+C \quad ($if $\quad n\neq-1)$
$\left[\begin{array}{ll}
g(x)=e^{x}, & u(x)=3e^{x}-1,\\
& u'(x)=3e^{x}
\end{array}\right]$
$=\displaystyle \frac{e^{x}}{3e^{x}}\cdot\frac{(3e^{x}-1)^{1/2}}{1/2}+C$
= $\displaystyle \frac{2}{3}\sqrt{3e^{x}-1}+C$
