integration of log of x plus square root of x square plus a square

int[log(x+sqrt(1+x^2))]/[sqrt(1+x^2)] dx

use integration by substitution twice.
first substitution,
let u = x + sqrt(1+x^2)
then du = [1 + x/sqrt(1+x^2)]dx
= [(sqrt(1+x^2) + x]/sqrt(1+x^2) dx
= [u/sqrt(1+x^2)]dx
du/u = dx/sqrt(1+x^2)

then int[ log(x+sqrt(1+x^2))/sqrt(1+x^2) dx]
= int[ log(x+sqrt(1+x^2) * dx/sqrt(1+x^2) ]
= int[ log(u) du/u ]

second substitution,
let v = log(u)
then dv = 1/u du = du/u

then int[ log(u) du/u ] = int( v dv)
= v^2/2 + C
= log(u)^2/2 + C.

hence,
int[ log(x+sqrt(1+x^2))/sqrt(1+x^2) dx]
= int (log(u) du/u)
= log(u)^2/2
= (log(x+sqrt(1+x^2)))^2/2
OR
= (sinh^(-1)(x))^2/2