Asymptotic Expansion in zero of $\frac{1}{\ln(1+x)}$

Question

On wolfram the expansion is:

$$\frac {1}{x} + \dfrac{1}{2} ...\,.$$

But I don't understand from where it outside comes the $\frac{1}{2}$

thanks

Answer 1

Note that, the function has a pole of order $1$ with residue equals $1$. Now, just write the function as

$$ \frac{1}{\ln(1+x)}=\frac{1}{x}+a_1+a_2 x+\dots \implies a_1= \lim_{x\to 0}\left(\frac{1}{\ln(1+x)}-\frac{1}{x} \right). $$