$\displaystyle f(x)=\frac{2x}{(x+1)^2(x^2+1)}=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}$

Find the seires expansion of $\displaystyle \log \left[ \frac{f(x)}{2x}\right]$ up to $\displaystyle [x^4]$

$\displaystyle \log\left[\frac{f(x)}{2x}\right]=\log\left[\frac{1}{(x+1)^2(x^2+1)}\right]$

I suppose then we would have $\displaystyle -2\log(x+1)-\log(x^2+1)$ but i don't know how to proceed after this.

Thanks