Answer
$$-\frac{2^{-(3x+2)}}{3\ln 2} +c $$
Work Step by Step
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
$$\int \left(\frac{1}{2}\right)^{3x+2}dx = \int 2^{-(3x+2)} dx\\
= -\frac{1}{3\ln 2} \int du=-\frac{1}{3\ln 2}u +c\\
=-\frac{2^{-(3x+2)}}{3\ln 2} +c. $$
