You also may use derivatives as alternative method in finding inverses, such that:

`(f^(-1)(x))' = 1/(f'(x))`

You need to evaluate `f'(x)` such that:

`f'(x) = ((2x-1)'(2x+1) - (2x-1)(2x+1)')/((2x+1)^2)`

`f'(x) = (2(2x + 1) - 2(2x - 1))/((2x+1)^2)`

`f'(x) = (4x + 2 - 4x + 2)/((2x+1)^2)`

`f'(x) = 4/((2x+1)^2) => 1/(f'(x)) = 1/(4/((2x+1)^2))`

`1/(f'(x)) = (2x+1)^2/4 => (f^(-1)(x))' = (2x+1)^2/4`

You need to integrate ` (f^(-1)(x))'` to evaluate `f^(-1)(x)` , such that:

`int (f^(-1)(x))' = int (2x+1)^2/4 dx`

You need to come up with the substitution, such that:

`2x + 1 = t => 2dx = dt => dx = (dt)/2`

Changing the variable, yields:

`int t^2/4*(dt)/2 = (1/8)t^3/3`

`int (2x+1)^2/4 dx = (2x + 1)^3/24 + c`

`int (2x+1)^2/4 dx = x^3/3 + x^2/4 + x/2 + 1/24 + c`

**Hence, evaluating the inverse of the function, using derivatives, yields **`f^(-1)(x) = x^3/3 + x^2/4 + x/2 + 1/24 + c.`

First, we'll write:

f(x)=(2x-1)/(2x+1) as y=(2x-1)/(2x+1)

Now, we'll solve this equation for x, multiplying both sides by (2x+1):

2xy+y = (2x-1)

We'll move all terms containing x, to the left side and all terms in y, to the right side:

2xy-2x = -1-y

We'll factorize by x:

x(2y-2) = -(1+y)

x=-(1+y)/2(y-1)

We'll multiply the denominator by -1 and we'll get:

x = (1+y)/2(1-y)

Now, we'll interchange x and y.

y = (1+x)/2(1-x)

So, the inverse function is:

[f(x)]^(-1) =(1+x)/2(1-x)