Answer
$\frac{2x(1-3x^2)}{(x^2+1)^5}$
Work Step by Step
Start with cancelling the common factor $(x^2+1)^3$, we have:
$\frac{(x^2+1)^4(2x)-x^2(4)(x^2+1)^3(2x)}{(x^2+1)^8}=\frac{(x^2+1)(2x)-x^2(4)(2x)}{(x^2+1)^5}=\frac{2x(1-3x^2)}{(x^2+1)^5}$
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