integrate between 0 to infinity [3/ (x²+1)]dx, where[ ] represents greatest integer function

When 3/(x^2 +1) < 1, greatest integer of the above = o Therefore, for x^2 + 1 < 3 => x > 2^(1/2), [3/(x^2 +1)] = o The integral can be written as integral of [3/(x^2 +1)] from 0 to sqrt 2. [3/(x^2+1)] = 1 for (1/sqrt 2 < x < sqrt 2) [3/(x^2+1)] = 2 for (0 < x < 1/sqrt2) [3/(x^2+1)] = 3 for (x=0) integral = (sqrt2 - (1/sqrt2)) + 2(1/sqrt2) = 3/sqrt2

so for 0 < x < 2^-1/2   : [3/ (x²+1)] = 2

and for 2^-1/2 < x < 2^1/2   : [3/ (x²+1)] = 1
and for 2^1/2 < x < infininty   : [3/ (x²+1)] = 0

so the final integral is :

 int. frm 0 to 2^-1/2: (2.dx) + int. frm 2^-1/2 to 2^1/2: (1.dx)

answer is:  3 / 2^1/2