We can simplify the expression by finding a common denominator: \begin{align*} \frac{1}{x-1}-\frac{1}{x-7}&>1\quad\Rightarrow\\ \frac{x-7}{(x-1)(x-7)}-\frac{x-1}{(x-1)(x-7)}&>1\quad\Rightarrow\\ \frac{-6}{x^2-8x+7}&>1.\end{align*}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\neq 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{align*} -6&>x^2-8x+7\quad\Rightarrow\\ 0&>x^2-8x+13.\end{align*}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{align*} -6&<x^2-8x+7\quad\Rightarrow\\ 0&<x^2-8x+13.\end{align*}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)} $.