Answer
$$\frac{1}{{\pi \ln 2}}\left( {{2^{\pi x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{2^{\pi x}}} dx \cr
& {\text{substitute }}u = \pi x \cr
& du = \pi dx \cr
& \frac{1}{\pi }du = dx \cr
& \int {{2^{\pi x}}} dx \cr
& = \int {{2^u}\left( {\frac{1}{\pi }du} \right)} \cr
& = \frac{1}{\pi }\int {{2^u}} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{\pi }\left( {\frac{{{2^u}}}{{\ln 2}}} \right) + C \cr
& = \frac{1}{{\pi \ln 2}}\left( {{2^u}} \right) + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \pi x \cr
& = \frac{1}{{\pi \ln 2}}\left( {{2^{\pi x}}} \right) + C \cr} $$
