jmc

We can simplify the expression by finding a common denominator: $$\begin{array}{rl}\frac{1}{x-1}-\frac{1}{x-7}& >1\Rightarrow \\ \frac{x-7}{(x-1)(x-7)}-\frac{x-1}{(x-1)(x-7)}& >1\Rightarrow \\ \frac{-6}{x^2-8x+7}& >1.\end{array}$$We would like to multiply both sides by $x^2-8x+7$, but we need to be careful: if $x^2-8x+7$ is negative, we will need to switch the inequality sign. We have two cases: $x^2-8x+7<0$ and $x^2-8x+7>0$. (Note that $x^2-8x+7\ne 0$ since it is in the denominator of a fraction.)

First let $x^2-8x+7>0$. Since the quadratic factors as $(x-7)(x-1)$, it changes sign at $x=7$ and $x=1$. Testing values reveals that the quadratic is positive for $x<1$ and $7<x$. Now since it is positive we can multiply both sides of the above inequality without changing the inequality sign, so we have $$\begin{array}{rl}-6& >x^2-8x+7\Rightarrow \\ 0& >x^2-8x+13.\end{array}$$The roots of the equation $x^2-8x+13$ occur at $$\frac{-(-8)\pm \sqrt{(-8{)}^2-4(1)(13)}}{2(1)}=\frac{8\pm \sqrt{12}}{2}=4\pm \sqrt{3}.$$Testing reveals that $x^2-8x+13<0$ when $x$ has a value between the roots, so $4-\sqrt{3}<x<4+\sqrt{3}$. However, we also have that either $x<1$ or $x>7$. Since $4-\sqrt{3}>1$ and $4+\sqrt{3}<7$, we actually have no values of $x$ which satisfy both inequalities.

Thus we must have $x^2-8x+7<0$. This occurs when $1<x<7$. When we cross-multiply, we must switch the inequality sign, so we have $$\begin{array}{rl}-6& <x^2-8x+7\Rightarrow \\ 0& <x^2-8x+13.\end{array}$$We already know the roots of the equation $x^2-8x+13$ are $4\pm \sqrt{3}$, so the sign of the quadratic changes there. We test to find that the quadratic is negative when $x<4-\sqrt{3}$ or $x>4+\sqrt{3}$. Combining this with the inequality $1<x<7$ gives us two intervals on which the inequality is satisfied: $\boxed{(1,4-\sqrt{3})\cup (4+\sqrt{3},7)}$.