Here comes annother gereral rule:
One always should have a rough idea about the function which is integrated.
In the case of 1/Log[x] there is a pole at x=1. So the integral has to be understood as "Cauchy principal value":
In[10]:= Integrate[1/Log[x], {x, 0, 2}, PrincipalValue -> True]
Out[10]= LogIntegral[2]
In[11]:= % // N
Out[11]= 1.04516
then it works perfect!
Cheers Henrik
In[13]:= $Version
Out[13]= "10.0 for Linux x86 (32-bit) (September 10, 2014)"
