Evaluate $I=\int_{\gamma (0,1)}\frac{zdz}{(z^2-4z+1)^2}$ using Taylor's Theorem I've shown that $f(z)=\frac{z}{(z^2-4z+1)^2}$ is holomorphic apart from at points $\alpha=2-\sqrt3$ and $\beta=2+\sqrt3$  and that the talyor coefficient of $g(z)=\frac{z}{(z-\beta)^2}$ centred at $\alpha$ is 
$$c_n=\frac{\alpha+n\beta}{(\beta-\alpha)^{n+2}}$$
Now I'm not sure how to use Taylor's theorem to evaluate the integral I, where $\gamma (0,1)$ is the unit circle centred at the origin
Thanks!
ADDITION
I can't seem to comment, but no, I don't think we've learnt residues. I have Cauchy's integral formula if that's similar?
Thank you so much for taking the time to answer my question. It's much much clearer now. I can't seem to upvote or select an answer though, any idea why this is?
 A: This answer assumes $\gamma(0,1)$ is $\gamma\colon [-\pi, \pi]\to \Bbb C, \theta \mapsto e^{i\theta}$.
The Residue theorem yields $$\displaystyle\int_\gamma f=2\pi i\left(\operatorname{Res}(f,\alpha)+\operatorname{Res}(f,\beta)\cdot 0\right)=2\pi i\operatorname{Res}(f,\alpha).$$
All that is left to find is $\operatorname{Res}(f,\alpha)$. You should know that $\operatorname{Res}(f,\alpha)$ is the coefficient of $(z-\alpha)^{-1}$ in the Laurent series of $f$ in a set like $\{z\in \Bbb C\colon 0<|z-\alpha|<r\}$, for some $r>0$.
You've already found, for all $z$ in the set above,  $\displaystyle \dfrac z{(z-\beta)^2}=\sum \limits_{n=0}^{+\infty}\left(c_n(z-\alpha)^n\right)$, thus
$$\dfrac{z}{(z^2-4z+1)^2}=\dfrac z{(z-\alpha)^2(z-\beta)^2}=\sum \limits_{n=0}^{+\infty}\left(c_n(z-\alpha)^{n-2}\right).$$
Therefore the coefficient of $(z-\alpha)^{-1}$of $f$ in its Laurent series is $c_1\color{grey}{=\dfrac{4}{(2\sqrt 3)^3}=\dfrac{1}{6\sqrt 3}}.$ Which agrees with Wolfram Alpha.
Finally $\displaystyle\int_\gamma f=2\pi i\cdot \dfrac{1}{6\sqrt 3}=\dfrac{\pi i}{3\sqrt 3}$.

In this part of the answer I'm mainly following theorem 5.2.1 in this notes.
Taylor's theorem (or perhaps just its proof, depending on the exact statement), says that in the series of $\varphi$ around $z_0$, $$\color{grey}{\varphi (z)=}\sum \limits_{n=0}^{+\infty}\left(a_n(z-z_0)^n\right),$$ for each $n\in \Bbb N_0$, $a_n$ is given by $$\displaystyle \color{grey}{a_n=}\dfrac 1{2\pi i}\int_{C_r}\dfrac{\varphi (z)}{(z-z_0)^{n+1}}\mathrm dz.$$
Setting $n=1,\varphi =g$ and $z_0=\alpha$ gives us $\displaystyle \int _{C_1}f=2\pi i\cdot c_1=\dfrac{\pi i}{3\sqrt 3}$.
