Evaluate the following integral:

let us assume![]()
let x = tany
differentiating both sides
dx = sec2 y dy
for x = ∞
![]()
For x = 0
0 = y
thus


(since sec2y – tan2y = 1)
.....equation 1
By property, we know that ![]()

….....equation 2
Adding equations 1 and 2, we get,

We know


since logm + logn = logmn
since tany = 1/coty
since log 1 = 0
Thus
2I = 0
I = 0
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