Answer
$\dfrac{2^{\pi+r}}{\ln 2} +C$
Work Step by Step
Calculate the anti-derivative.
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $ \int a^x dx=\dfrac{1}{\ln a} a^x+c$
where $c$ is a constant of proportionality.
Then $\int 2^{\pi+r} dr=(\dfrac{1}{\ln 2})(2^{\pi+r})+c$
Thus, $= \dfrac{2^{\pi+r}}{\ln 2} +c$
