Answer
$\frac{4}{(x^2+4)\sqrt{x^2+4}}$
Work Step by Step
First, re-write to make it in radical form:
$\frac{\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}}{x^2+4}$
Then, make both denominators of the top fractions the same:
$\frac{\frac{\sqrt{x^2+4}\cdot\sqrt{x^2+4}}{\sqrt{x^2+4}}-\frac{{x^2}}{\sqrt{x^2+4}}}{x^2+4}$
Because of the law of radicals $\sqrt[n] a \cdot \sqrt[n] b=\sqrt[n] {ab}$:
$\frac{\frac{\sqrt{(x^2+4)^2}}{\sqrt{x^2+4}}-\frac{x^2}{\sqrt{x^2+4}}}{x^2+4}$
Simplify further:
$\frac{\frac{x^2+4-x^2}{\sqrt{x^2+4}}}{x^2+4}=\frac{4}{\sqrt{x^2+4}}\cdot\frac{1}{x^2+4}=\frac{4}{(x^2+4)\sqrt{x^2+4}}$
