<s>[INST] Show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$ when $x>1$ 
If $x>1$ show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$

I know from binomial expansion that $(1+x)$ will produce a divergent series in the form of $1-x+x^2-x^3+\cdots$ but I don't know how to apply that in this situation. 
Do I just need to integrate $\ln(1-x+x^2+\cdots$)? If so, why integrate? What role does integration have here?
 [/INST] Let
$$\begin{align}
f(x)&=\ln(1+x)-\ln x\\
&=\ln\left(1+\frac{1}{x}\right)
\end{align}
$$
By differenting w.r.t. $x$ and writing as sum of infinite GP
$$\begin{align}f'(x)&=\frac{1}{1+\frac1x}\cdot\left(\frac{-1}{x^2}\right)\\
&=-\frac{1}{x^2}\left(1-\frac1x+\frac{1}{x^2}-\cdots\right)\\
f'(x)&=-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\cdots\\
\end{align}$$
By integrating
$$\begin{align}
f(x)&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\
\ln(1+x)-\ln x&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\
\end{align}$$

$$\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\cdots$$

</s>