1. The problem is to evaluate the double integral $$\int_0^1 \int_1^2 x(x+1) \, dy \, dx.$$ 2. The integral is over $y$ from 1 to 2 and $x$ from 0 to 1. The integrand is $x(x+1)$, which does not depend on $y$. 3. Since the integrand does not depend on $y$, the inner integral with respect to $y$ is simply the integrand times the length of the $y$-interval: $$\int_1^2 x(x+1) \, dy = x(x+1)(2-1) = x(x+1).$$ 4. Now the problem reduces to a single integral: $$\int_0^1 x(x+1) \, dx = \int_0^1 (x^2 + x) \, dx.$$ 5. Integrate term-by-term: $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3},$$ $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}.$$ 6. Add the results: $$\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}.$$ 7. Therefore, the value of the double integral is $$\boxed{\frac{5}{6}}.$$