jmc

First, let's deal with $|x|+1>7$. Subtracting 1 from both sides gives $|x|>6$, so the integers that satisfy $|x|+1>7$ are those greater than 6 and those less than $-6$. Since the inequality is strict ($>$, not $\ge$), $x$ cannot be 6 or $-6$.

Next, we consider $|x+1|\le 7$. Writing this as $|x-(-1)|\le 7$, we see that $x$ must be within $7$ of $-1$ on the number line, which means it must be one of the integers from $-8$ to 6. Since the inequality is nonstrict ($\le$, not $<$), $x$ can be $-8$ or 6.

The only integers that satisfy both inequalities are $-8$ and $-7$, and their sum is $\boxed{-15}$.