$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z} $$
Since -2 lies in $|z+1| = 2$, Cauchy theorem gives that
$$\frac{1}{2\pi i} \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{z-(-2)} = \frac{-2}{2-(-2)} = -\frac{1}{2}$$
then $$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = -\pi i$$
However, I checked the answer of textbook and it says the answer is $2\pi i$, I am confused about where i did wrong.
