Ok, firstly, you have x-1, x+1, and x^2-1 as the denominator in your question. Thus, I will take it as the question implicitly assumes that x != 1 or -1. This is actually pretty important.
Let's combine the fraction on the right into a single fraction,
x/(x-1) + 4/(x+1) = (x(x+1))/((x-1)(x+1)) + (4(x-1))/((x-1)(x+1)) = (x^2 + x + 4x - 4)/(x^2-1) = (x^2 + 5x -4)/(x^2 -1)
Here, note that (x-1)(x+1) = x^2 - 1 from difference of two squares.
We have:
(x^2 + 5x -4)/(x^2 -1) = (4x-2)/(x^2-1)
Cancel out the denominator (multiply both sides by x^2-1),
x^2 + 5x -4 = 4x-2
Please note that this step is only possible due to our assumption at the start. Cancelling (x^2-1)/(x^2-1) = 1 is only valid for x^2-1 != 0.
x^2 + x -2 = 0
We can factorise this quadratic equation:
x^2 + x - 2 = (x - 1)(x + 2) = 0
And thus, x = 1, or x = -2.
But we're not done yet. This is the solution to the quadratic equation, but not the equation in the question.
In this case, x = 1 is an extraneous solution, which is an extra solution that is generated by the way we solve our problem, but is not an actual solution.
So, we reject x = 1, from our assumption earlier.
Therefore, x = -2.