Answer
$\frac{1}{x^2\sqrt{x^2-1}}$
Work Step by Step
First, re-write to make it in radical form:
$\frac{\frac{x^2}{\sqrt{x^2-1}}-\sqrt{x^2-1}}{x^2}$
Then, make both denominators of the top fraction the same:
$\frac{\frac{x^2}{\sqrt{x^2-1}}-\frac{\sqrt{x^2-1}\cdot\sqrt{x^2-1}}{\sqrt{x^2-1}}}{x^2}$
Because of the law of radicals $\sqrt[n] a \cdot \sqrt[n] b=\sqrt[n] {ab}$:
$\frac{\frac{x^2}{\sqrt{x^2-1}}-\frac{\sqrt{(x^2-1)^2}}{\sqrt{x^2-1}}}{x^2}$
Simplify further:
$\frac{\frac{x^2-(x^2-1)}{\sqrt{x^2-1}}}{x^2}=\frac{1}{\sqrt{x^2-1}}\cdot\frac{1}{x^2}=\frac{1}{x^2\sqrt{x^2-1}}$
