What is the sum of all integer solutions to $1<(x-2)^2<25$?

$ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 $
1. Start with the given inequality:
   \[
   1 < (x-2)^2 < 25
   \]

2. To solve this, we need to consider the square root of the inequality. Since the square root function produces both positive and negative roots, we split the inequality into two parts:
   \[
   \sqrt{1} < |x-2| < \sqrt{25}
   \]
   Simplifying the square roots, we get:
   \[
   1 < |x-2| < 5
   \]

3. The absolute value inequality \(1 < |x-2| < 5\) can be split into two separate inequalities:
   \[
   1 < x-2 < 5 \quad \text{or} \quad -5 < x-2 < -1
   \]

4. Solve each inequality separately:
   - For \(1 < x-2 < 5\):
     \[
     1 < x-2 \implies x > 3
     \]
     \[
     x-2 < 5 \implies x < 7
     \]
     Combining these, we get:
     \[
     3 < x < 7
     \]

   - For \(-5 < x-2 < -1\):
     \[
     -5 < x-2 \implies x > -3
     \]
     \[
     x-2 < -1 \implies x < 1
     \]
     Combining these, we get:
     \[
     -3 < x < 1
     \]

5. Now, we find the integer solutions for each range:
   - For \(3 < x < 7\), the integer solutions are \(4, 5, 6\).
   - For \(-3 < x < 1\), the integer solutions are \(-2, -1, 0\).

6. Sum all the integer solutions:
   \[
   4 + 5 + 6 + (-2) + (-1) + 0 = 12
   \]

The final answer is \(\boxed{12}\).