What is the antiderivative of the function y, given by y=x^2/(x^2-1)?

To determine the antiderivative of the function, we'll have to
calculate the indefinite integral of the given function.

Int x^2
dx/(x^2-1) = Int (x^2 - 1)dx/(x^2 - 1) + Int dx/(x^2 - 1)

Int x^2
dx/(x^2-1) = Int dx + Int dx/(x^2 - 1)

We notice that the
denominator x^2 - 1 is a difference of 2 squares:

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

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

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

1 = Ax + A + Bx - B

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

Comparing, we'll get:

A+B =
0

A-B = 1

-2B = 1 => B = -1/2
=> A = 1/2

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

We'll integrate both
sides:

Int dx/(x^2 - 1) = Int dx/2(x-1) - Int
dx/2(x+1)

Int dx/(x^2 - 1) = (1/2)*[ln|x-1| - ln|x+1|] +
C

Int dx/(x^2 - 1) = (1/2)*[ln|x-1|/|x+1|] +
C

Int dx/(x^2 - 1) = ln sqrt[|x-1|/|x+1|] +
C

The antiderivative of the given function is: Int x^2
dx/(x^2-1) = x + ln sqrt[|x-1|/|x+1|] + C