Let $\{a_n\}_{n\geq 1}$ be a sequence defined by $a_n=\int_0^1 x^2(1-x)^ndx$.
Find the real value of $c$ such that $\sum_{n=1}^{\infty} (n+c)(a_n-a_{n+1})=2.$
1. First, we start with the given sequence \( a_n = \int_0^1 x^2 (1-x)^n \, dx \).
2. We need to find the value of \( c \) such that \( \sum_{n=1}^{\infty} (n+c)(a_n - a_{n+1}) = 2 \).

3. To proceed, we need to understand the behavior of \( a_n \) and \( a_{n+1} \). Let's compute \( a_n - a_{n+1} \):
   \[
   a_n = \int_0^1 x^2 (1-x)^n \, dx
   \]
   \[
   a_{n+1} = \int_0^1 x^2 (1-x)^{n+1} \, dx
   \]
   \[
   a_n - a_{n+1} = \int_0^1 x^2 (1-x)^n \, dx - \int_0^1 x^2 (1-x)^{n+1} \, dx
   \]
   \[
   = \int_0^1 x^2 (1-x)^n \left(1 - (1-x)\right) \, dx
   \]
   \[
   = \int_0^1 x^2 (1-x)^n x \, dx
   \]
   \[
   = \int_0^1 x^3 (1-x)^n \, dx
   \]

4. Now, we need to evaluate the sum \( \sum_{n=1}^{\infty} (n+c)(a_n - a_{n+1}) \):
   \[
   \sum_{n=1}^{\infty} (n+c) \int_0^1 x^3 (1-x)^n \, dx
   \]

5. We can interchange the sum and the integral (by Fubini's theorem):
   \[
   \int_0^1 x^3 \sum_{n=1}^{\infty} (n+c) (1-x)^n \, dx
   \]

6. The inner sum is a geometric series. We need to find the sum of \( \sum_{n=1}^{\infty} (n+c) (1-x)^n \):
   \[
   \sum_{n=1}^{\infty} n (1-x)^n + c \sum_{n=1}^{\infty} (1-x)^n
   \]

7. The first sum is:
   \[
   \sum_{n=1}^{\infty} n (1-x)^n = \frac{(1-x)}{(1-(1-x))^2} = \frac{(1-x)}{x^2}
   \]

8. The second sum is:
   \[
   \sum_{n=1}^{\infty} (1-x)^n = \frac{(1-x)}{x}
   \]

9. Combining these results:
   \[
   \sum_{n=1}^{\infty} (n+c) (1-x)^n = \frac{(1-x)}{x^2} + c \frac{(1-x)}{x}
   \]

10. Now, we integrate:
    \[
    \int_0^1 x^3 \left( \frac{(1-x)}{x^2} + c \frac{(1-x)}{x} \right) \, dx
    \]
    \[
    = \int_0^1 \left( x (1-x) + c x^2 (1-x) \right) \, dx
    \]

11. Split the integral:
    \[
    \int_0^1 x (1-x) \, dx + c \int_0^1 x^2 (1-x) \, dx
    \]

12. Evaluate each integral separately:
    \[
    \int_0^1 x (1-x) \, dx = \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}
    \]
    \[
    \int_0^1 x^2 (1-x) \, dx = \int_0^1 (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}
    \]

13. Combine the results:
    \[
    \frac{1}{6} + c \cdot \frac{1}{12} = 2
    \]

14. Solve for \( c \):
    \[
    \frac{1}{6} + \frac{c}{12} = 2
    \]
    \[
    \frac{c}{12} = 2 - \frac{1}{6}
    \]
    \[
    \frac{c}{12} = \frac{12}{6} - \frac{1}{6}
    \]
    \[
    \frac{c}{12} = \frac{11}{6}
    \]
    \[
    c = \frac{11}{6} \cdot 12
    \]
    \[
    c = 22
    \]

The final answer is \( \boxed{22} \).