Answer:
[tex]\frac{12x}{x^4-1}[/tex]
Explanation:
Given
[tex](\frac{x+1}{x-1} - \frac{x-1}{x+1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
Required
Solve:
Take L.C.M of the first two fractions
[tex](\frac{(x+1)(x+1)-(x-1)(x-1)}{(x-1)(x+1)} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{(x+1)(x+1)-(x-1)(x-1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{x^2+2x+1-(x^2-2x+1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{x^2+2x+1-x^2+2x-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
Collect Like Terms
[tex](\frac{x^2-x^2+2x+2x+1-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{4x}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}[/tex]
Solve the expression in bracket
[tex](\frac{4x(x^2+1) - 4x(x^2-1)}{(x^2-1)(x^2+1)} ) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{4x(x^2+1) - 4x(x^2-1)}{x^4-1} ) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{4x^3+4x - 4x^3+4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}[/tex]
[tex](\frac{4x^3- 4x^3+4x +4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}[/tex]
[tex]\frac{8x}{x^4-1} + \frac{4x}{x^4 - 1}[/tex]
Take LCM
[tex]\frac{8x+4x}{x^4-1}[/tex]
[tex]\frac{12x}{x^4-1}[/tex]
The expression can not be further simplified
