What is the inverse of f(x) = (x^2 +1) / (1- x^2)?

The inverse function of f(x) is defined as the function
which when applied to the result of f(x) gives x.

We have
f(x) = (x^2 +1) / (1- x^2), let this be equal to
y

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

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

=> y – y x^2 = x^2
+1

=> y -1 = yx^2 +
x^2

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

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

=> x = sqrt [(y-1) /
(y+1)]

Interchange y and x

y =
sqrt [(x-1) / (x+1)]

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

Therefore the inverse function of f(x)
= (x^2 +1) / (1- x^2) is f^-1(x) = sqrt [(x-1) /
(x+1)]

To verify: f(x) = (x^2 +1) / (1-
x^2)

substitute this in sqrt [(x-1) /
(x+1)]

=> sqrt [{(x^2 +1) / (1- x^2) -1} / { (x^2
+1) / (1- x^2)} +1]

=> sqrt[ (x^2 +1 – 1 + x^2) /
x^2 + 1 + 1 – x^2]

=> sqrt [ 2x^2
/2]

=> sqrt
x^2

=> x