Answer
$\dfrac{(x+1)(x-1)}{(x^2+1)^2}$
Work Step by Step
Use the distributive property.
$\dfrac{x\cdot 2x-(x^2+1)\cdot 1}{(x^2+1)^2}\\
\\=\dfrac{2x^2-(x^2+1)}{(x^2+1)^2}\\
\\=\dfrac{2x^2-x^2-1}{(x^2+1)^2}$
Simplify.
$=\dfrac{x^2-1}{(x^2+1)^2}\\$
$\\=\dfrac{x^2-1^2}{(x^2+1)^2}$
Use special rule $a^2-b^2=(a+b)(a-b)$.
$=\dfrac{(x+1)(x-1)}{(x^2+1)^2}$
