Then
Int sqrt(1 + x^2) d x
= Int (cosh t)^2 dt
= (1/2) Int [1 + cosh(2t)] d t
=(1/2) [t + (sinh t)(cosh t)]
=(1/2) [t + x sqrt(1 + x^2)]
To express t in term of x, we rewrite x = sinh t = [e^t - e^{-t}]/2 as
(e^t)^2 - 2x (e^t) - 1 = 0.
The quadratic formula then gives e^t = x + sqrt(1 + x^2), which yields
t = ln[x + sqrt(1 + x^2)].
Finally, Int sqrt(1 + x^2) d x = (1/2) {ln[x + sqrt(1 + x^2)] + x sqrt(1 + x^2)}
