Finding $f(z)=\log(\frac{(z+1)^2}{(z^2+4)})$ Laurent series

Determine the Laurent series of the function $$f(z)=\log(\frac{(z+1)^2}{(z^2+4)})$$ on the set $$A=\{z|\:2<|z|\}$$.

I thought of using the following expansion: $$\log(1+z)=\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}z^n }{n},|z|<1$$

Using the logarithm properties I tried to arrange the expression so that I could apply the formula: $$\log(\frac{(z+1)^2}{(z^2+4)})=\log z^3+2\log(1+\frac{1}{z})-\log(1+\frac{4}{z^2})$$

However I do not know what should I do regarding $$\log(z^3)$$.

The solution provided in the book is: $$\sum_\limits{-\infty}^{-1}\frac{1-(-1)^{n}4^{-n} }{n}z^{2n}-\sum_\limits{-\infty}^{0}\frac{2}{2n-1}z^{2n-1}$$

Question:

How should I solve the question?

Check your work. There should be no $$\log z$$ terms involved.
Note that the fraction is homogeneous [which also implies that you could avoid $$\log z$$ terms completely], so we do the following: $$f (z) = \log \left(\frac {(1 + z^{-1})^2} { 1 + 4z^{-2}}\right) = 2\log (1 + z^{-1}) - \log (1+4z^{-2}),$$ since $$|z| > 2$$, $$|z^{-1}| < 1/2 < 1, |4z^{-2}|< 1$$, then $$f(z) = 2 \sum_1^{+\infty} \frac {(-1)^{n-1}}n z^{-n} - \sum_1^{+\infty} \frac {(-1)^{n-1}4^n }n z^{-2n}$$ or $$f(z) = 2 \sum_{-\infty}^1 \frac {(-1)^n}n z^n - \sum_{-\infty}^1 \frac {(-1)^n4^{-n}}n z^{2n}.$$ Now collect the odd-indexed terms and even-indexed terms respectively, then you would reach the answer provided.
Since$$\log(z+1)=z-\frac{z^2}2+\frac{z^3}3-\frac{z^4}4+\cdots,$$you know that$$\log\left((z+1)^2\right)=2\log(z+1)=2z-z^2+\frac{2z^3}3-\frac{2z^4}4+\cdots,$$and that$$\log(4z^2+1)=4z^2-\frac{4^2z^4}2+\frac{4^3z^6}3-\frac{4^4z^8}4+\cdots$$and so$$\log\left(\frac{(z+1)^2}{4z^2+1}\right)=2z-5z^2+\frac23z^3+\frac{15}2z^4+\cdots$$Therefore, if $$\lvert z\rvert>2$$, $$\left\lvert\frac1z\right\rvert<\frac12$$ and\begin{align}\log\left(\frac{(z+1)^2}{z^2+4}\right)&=\log\left(\frac{(\frac1z+1)^2}{4\frac1{z^2}+1}\right)\\&=\frac2z-\frac5{z^2}+\frac2{3z^3}+\frac{15}{2z^4}+\cdots\end{align}