Answer
$\displaystyle \frac{1}{6}\ln|1+2e^{3x}|+C$
Work Step by Step
$\displaystyle \int\frac{e^{3x}}{1+2e^{3x}}dx=\int e^{3x}(1+2e^{3x})^{-1}dx=$
Shortcut to apply:
$\displaystyle \quad \int g\cdot u^{-1}dx=\frac{g}{u'}\ln|u|+C$
$\left[\begin{array}{ll}
g(x)=e^{3x}, & u(x)=1+2e^{3x}\\
& u'(x)=2e^{3x}\cdot 3=6e^{3x}
\end{array}\right]$
$=\displaystyle \frac{e^{3x}}{6e^{3x}}\ln|u|+C$
= $\displaystyle \frac{1}{6}\ln|1+2e^{3x}|+C$
