## Calculus (3rd Edition)

$$-\frac{2^{-(3x+2)}}{3\ln 2} +c$$
We have $$\int \left(\frac{1}{2}\right)^{3x+2}dx = \int 2^{-(3x+2)} dx.$$ Let $u=2^{-(3x+2)}$, then $du=-(3\ln 2) \ 2^{-(3x+2)} dx$ and hence $$\int \left(\frac{1}{2}\right)^{3x+2}dx = \int 2^{-(3x+2)} dx\\ = -\frac{1}{3\ln 2} \int du=-\frac{1}{3\ln 2}u +c\\ =-\frac{2^{-(3x+2)}}{3\ln 2} +c.$$