Find the number of integer solutions of x.

We will consider different cases in which the expressions inside the absolute value signs will be negative or positive.

Case 1: $x\le1$

Both $(x-1)$ and $(x-6)$ will be negative/zero. The absolute values will negate these expressions to make them positive. We end up with the inequality:

$-x+6-x+1\le13$

Solving this inequality yields: $x\ge-3$ . The intersection of $x\le1$ and $x\ge-3$ is:

$-3\le x\le1$

Case 2: $1<x\le6$

$(x-1)$ will be positive while $(x-6)$ will be negative/zero. The absolute value sign will do nothing to $(x-1)$ and the absolute value sign will negate $(x-6)$ in order to make it positive. We end up with the inequality:

$-x+6+x-1\le13$

Solving this yields $5\le13$ , which is true regardless of the value of $x$ . Therefore, the inequality is true for the entire interval of this case.

Case 3: $x>6$

Both $(x-1)$ and $(x-6)$ will be positive/zero. The absolute value signs will do nothing to either expression. We end up with the inequality:

$x-6+x-1\le13$

Solving this yields $x\le10$ . The intersection of $x>6$ and $x\le10$ is $6<x\le10$

The union of the solutions of all three cases is: $-3\le x\le10$ , and there are $\boxed{14}$ integers in this interval.