Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises - Page 481: 59

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

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