1. **State the problem:** We need to evaluate the double integral $$\int_0^1 \int_0^1 v (u + v^2)^4 \, du \, dv.$$\n\n2. **Understand the integral:** The integral is over $u$ from 0 to 1 and $v$ from 0 to 1. The integrand is $v (u + v^2)^4$.\n\n3. **Integrate with respect to $u$ first:** Treat $v$ as a constant during this integration.\n\n$$\int_0^1 v (u + v^2)^4 \, du = v \int_0^1 (u + v^2)^4 \, du.$$\n\n4. **Use substitution for the inner integral:** Let $w = u + v^2$, then $dw = du$. When $u=0$, $w = v^2$. When $u=1$, $w = 1 + v^2$.\n\nSo,\n$$\int_0^1 (u + v^2)^4 \, du = \int_{v^2}^{1+v^2} w^4 \, dw = \left[ \frac{w^5}{5} \right]_{v^2}^{1+v^2} = \frac{(1+v^2)^5 - (v^2)^5}{5}.$$\n\n5. **Substitute back:**\n$$\int_0^1 v (u + v^2)^4 \, du = v \cdot \frac{(1+v^2)^5 - v^{10}}{5} = \frac{v}{5} \left((1+v^2)^5 - v^{10}\right).$$\n\n6. **Now integrate with respect to $v$ from 0 to 1:**\n$$\int_0^1 \frac{v}{5} \left((1+v^2)^5 - v^{10}\right) dv = \frac{1}{5} \int_0^1 v (1+v^2)^5 dv - \frac{1}{5} \int_0^1 v^{11} dv.$$\n\n7. **Evaluate each integral separately:**\n- For $$\int_0^1 v (1+v^2)^5 dv,$$ use substitution $t = 1 + v^2$, so $dt = 2v dv$, or $v dv = \frac{dt}{2}$. When $v=0$, $t=1$, when $v=1$, $t=2$.\n\nThus,\n$$\int_0^1 v (1+v^2)^5 dv = \int_1^2 t^5 \frac{dt}{2} = \frac{1}{2} \int_1^2 t^5 dt = \frac{1}{2} \left[ \frac{t^6}{6} \right]_1^2 = \frac{1}{12} (2^6 - 1^6) = \frac{1}{12} (64 - 1) = \frac{63}{12} = \frac{21}{4}.$$\n\n- For $$\int_0^1 v^{11} dv = \left[ \frac{v^{12}}{12} \right]_0^1 = \frac{1}{12}.$$\n\n8. **Combine results:**\n$$\frac{1}{5} \cdot \frac{21}{4} - \frac{1}{5} \cdot \frac{1}{12} = \frac{21}{20} - \frac{1}{60} = \frac{63}{60} - \frac{1}{60} = \frac{62}{60} = \frac{31}{30}.$$\n\n**Final answer:** $$\boxed{\frac{31}{30}}.$$