1. **State the problem:** Evaluate the integral $$\int 6 \cdot 2^{3x} \, dx$$. 2. **Recall the formula:** The integral of an exponential function $$a^{kx}$$ with respect to $$x$$ is given by $$\int a^{kx} \, dx = \frac{a^{kx}}{k \ln(a)} + C$$, where $$a > 0$$ and $$a \neq 1$$. 3. **Apply the formula:** Here, $$a = 2$$ and $$k = 3$$, so $$\int 6 \cdot 2^{3x} \, dx = 6 \int 2^{3x} \, dx = 6 \cdot \frac{2^{3x}}{3 \ln(2)} + C$$. 4. **Simplify the constant factor:** $$6 \cdot \frac{1}{3} = \cancel{6} \cdot \frac{1}{\cancel{3}} = 2$$, so the integral becomes $$\frac{2 \cdot 2^{3x}}{\ln(2)} + C = \frac{2^{3x + 1}}{\ln(2)} + C$$. 5. **Final answer:** $$\int 6 \cdot 2^{3x} \, dx = \frac{2^{3x + 1}}{\ln(2)} + C$$. This means the antiderivative of $$6 \cdot 2^{3x}$$ is $$\frac{2^{3x + 1}}{\ln(2)}$$ plus the constant of integration.