Answer
$\frac{1-3x^2}{2(1+x^2)^2(\sqrt{x})}$
Work Step by Step
First,, make both denominators of the top fraction the same:
$\frac{\frac{1+x^2}{2\sqrt{x}}-\frac{2x\sqrt{x}\cdot2\sqrt{x}}{2\sqrt{x}}}{(1+x^2)^2}$
Because of the law of radicals $\sqrt[n] a \cdot \sqrt[n] b=\sqrt[n] {ab}$:
$\frac{\frac{1+x^2}{2\sqrt{x}}-\frac{2x\cdot2\sqrt{x^2}}{2\sqrt{x}}}{(1+x^2)^2}$
Simplify further:
$\frac{\frac{1+x^2-4x^2}{2\sqrt{x}}}{(1+x^2)^2}=\frac{1-3x^2}{2\sqrt{x}}\cdot\frac{1}{(1+x^2)^2}=\frac{1-3x^2}{2(1+x^2)^2(\sqrt{x})}$
