#### Answer

$$ - \frac{{{3^{ - 2x}}}}{{2\ln 3}} + C$$

#### Work Step by Step

$$\eqalign{
& \int {{3^{ - 2x}}} dx \cr
& {\text{substitute }}u = - 2x,{\text{ }}du = - 2dx \cr
& = - \frac{1}{2}\int {{3^u}} du \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{2}\left( {\frac{{{3^u}}}{{\ln 3}}} \right) + C \cr
& {\text{with }}u = - 2x \cr
& = - \frac{1}{2}\left( {\frac{{{3^{ - 2x}}}}{{\ln 3}}} \right) + C \cr
& = - \frac{{{3^{ - 2x}}}}{{2\ln 3}} + C \cr} $$