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integrate between 0 to infinity [3/ (x2+1)]dx, where[ ] represents greatest integer function
so for 0 < x < 2-1/2 : [3/ (x2+1)] = 2
and for 2-1/2 < x < 21/2 : [3/ (x2+1)] = 1 and for 21/2 < 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 21/2: (1.dx)
answer is: 3 / 21/2
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