Value of $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz$ $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz=-2\pi i$.
Am  I correct, I used cauchy integral formula


Since $z^2-4=(z-2)(z+2)$ and $z=-2$ is the only zero inside $|z+1|=2$. Assuming $|z+1|=2$ is positively oriented, Cauchy's Integral Formula gives 
$$
\int_{|z+1|=2}\frac{z^2}{4-z^2}dz =- \int_{|z+1|=2}\frac{z^2}{z^2-4}dz = -\int_{|z+1|=2}\frac{z^2}{(z-2)(z+2)}dz =- 2 \pi i \left( \frac{z^2}{z-2} \right)_{z=-2} = 2\pi i
$$
Hence your result is not the correct one! If $|z+1|=2$ is negatively oriented, then you have the correct result!
