The expression is $$\frac{1 + x}{1 - x} -\frac{1 - x}{1 + x} + \frac{4x}{1 + x^2} + \frac{8x^3}{1 - x^4}$$ is equal to:

Correct option is D

​Given:

​$$\frac{1+x}{1-x} - \frac{1-x}{1+x} + \frac{4x}{1+x^2} + \frac{8x^3}{1-x^4}$$​​

Formula Used:

​$$1-x^4 = (1-x^2)(1+x^2)$$​​

Solution: 

​​$$\frac{1+x}{1-x} - \frac{1-x}{1+x} + \frac{4x}{1+x^2} + \frac{8x^3}{1-x^4} \\ \ \\ = \frac{(1+x)^2-(1-x)^2}{(1+x)(1-x)} + \frac{4x}{1+x^2} + \frac{8x^3}{1-x^4} \\ \ \\ = \frac{4x}{1-x^2}+\frac{4x}{1+x^2} + \frac{8x^3}{1-x^4} \\ \ \\ = \frac{4x(1+x^2+1-x^2)}{(1-x^2)(1+x^2)} + \frac{8x^3}{1-x^4} \\ \ \\ = \frac{8x}{(1-x^2)(1+x^2)} + \frac{8x^3}{(1-x^2)(1+x^2)}\\ \ \\ =\frac{8x(1+x^2)}{(1+x^2)(1-x^2)} \\ \ \\ = \frac{8x}{1-x^2}$$​​