What is \frac{8x}{1-x^{4}}-\frac{4x}{x^{2}+1}+\frac{x+1}{x-1}-\frac{x-1}{x+1} equal to?

Explanation

First simplify \frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{(x+1)^2 - (x-1)^2}{x^2-1} = \frac{4x}{x^2-1} = -\frac{4x}{1-x^2}. Next, evaluate -\frac{4x}{1-x^2} - \frac{4x}{1+x^2} = -4x\left(\frac{1+x^2+1-x^2}{1-x^4}\right) = -\frac{8x}{1-x^4}. Finally, adding this to the first term \frac{8x}{1-x^4} results in 0.

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