Write 1+i=z in exponential form:
|z|=sqrt(1^2+1^2)=sqrt(2)
tan(phi)=1/1, phi=arctan(1)+2 * k * Pi=0.7854+2 * k * Pi
z=1+i=sqrt(2) * e^(i * [0.7854+2 * k * Pi]
x = ln(1+i) = ln{  sqrt(2) * e^(i * [0.7854+2 * k * Pi] } = ln(sqrt(2)) + i * [0.7854+2* k * Pi]
x = 0.3466 + i(0.7854+ 2 * k * Pi)
With k in N