You should remember that the antiderivative is a function F(x), having the property `F'(x) = f(x). ` Since the reverse operation to differentiation is integration, you need to integrate the given function to find antiderivative F(x) such that:

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

Using the property of integral yields:

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

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

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

You need to use partial fraction decomposition to write the fraction `1/((x-1)(x+1))` as a sum of simpler fractions such that:

`1/((x-1)(x+1)) = A/(x-1) +B/(x+1)`

`1 =Ax + A + Bx- B `

`1 = x(A+B) + A-B`

Equating the coefficients of like powers yields:

`A+B = 0 = gt A=-B`

`A-B = 1 =gt -B-B = 1 =gt -2B = 1 =gt B = -1/2 =gt A = 1/2`

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

Hence, evaluating the integral yields:

`int 1/((x-1)(x+1)) dx = int (dx)/(2(x-1)) -int (dx)/(2(x+1))`

`int 1/((x-1)(x+1)) dx = (1/2) ln|x-1| - (1/2)ln |x+1| + c`

`int 1/((x-1)(x+1)) dx = (1/2) ln |(x-1)/(x+1)| + c`

`int 1/((x-1)(x+1)) dx = ln |(x-1)/(x+1)|^(1/2) + c`

`int 1/((x-1)(x+1)) dx = ln sqrt ((x-1)/(x+1)) + c`

`int x^2/(x^2-1) dx = x +ln sqrt ((x-1)/(x+1)) + c`

**Hence, evaluating the antiderivative of the function yields `F(x) = x + ln sqrt ((x-1)/(x+1)) + c.` **