1. **State the problem:** We need to evaluate the definite integral $$\int_0^1 \left(0.16(\log(x))^2 + 0.704\log(x) + 0.7744\right) dx.$$ 2. **Recall useful formulas:** For $a > -1$, the integral $$\int_0^1 (\log x)^a dx = (-1)^a a!$$ if $a$ is an integer, but more generally, $$\int_0^1 (\log x)^n dx = (-1)^n n!$$ for integer $n \geq 0$. Specifically, $$\int_0^1 \log(x) dx = -1$$ $$\int_0^1 (\log(x))^2 dx = 2$$ $$\int_0^1 dx = 1.$$ 3. **Break the integral into parts:** $$\int_0^1 0.16(\log(x))^2 dx + \int_0^1 0.704 \log(x) dx + \int_0^1 0.7744 dx.$$ 4. **Evaluate each integral:** - $$\int_0^1 (\log(x))^2 dx = 2,$$ so $$0.16 \times 2 = 0.32.$$ - $$\int_0^1 \log(x) dx = -1,$$ so $$0.704 \times (-1) = -0.704.$$ - $$\int_0^1 0.7744 dx = 0.7744 \times 1 = 0.7744.$$ 5. **Sum the results:** $$0.32 - 0.704 + 0.7744 = 0.3904.$$ **Final answer:** $$\boxed{0.3904}.$$