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