We have
\begin{align} \mathcal{J}=\int_{\partial \mathcal{D}(0,2)} \dfrac{2}{z^2+1}dz. \end{align}
Using the Cauchy Theorem we can write \begin{align} \int\dfrac{2}{(z-i)(z+i)}dz \end{align} we have 2 simple poles.
For the first pole, $z=-i$ write \begin{align} \dfrac{\Phi(z)}{(z+i)^1} \mapsto \mathrm{Res}(f)=\dfrac{\Phi^{(0)}(-i)}{(1-1)!} = \dfrac{2}{-2i}=i \end{align} and the second, $z=i$, \begin{align} \dfrac{\Phi(z)}{(z-i)^1} \mapsto \mathrm{Res}(f)=\dfrac{\Phi^{(0)}(i)}{(1-1)!} = \dfrac{2}{2i}=-i \end{align} and by the residue theorem \begin{align} \mathcal{J} = 2\pi i \sum (\text{residues)} = 2\pi i(i-i)=0. \end{align}
However, going for the direct parameterization, \begin{align} \gamma(t) = 2e^{it}, t \in [0,2\pi]\\ \gamma'(t) = 2i e^{it} \end{align} then we have \begin{align} \int_{0}^{2\pi} \dfrac{4ie^{it}}{4e^{2it}+1}dt = \ldots = [2\arctan(2e^{it})]_{0}^{2\pi} = \dfrac{\pi}{2} \end{align} where did I go wrong?
There is a dirty trick showing that the integral is indeed equal $0.$ We have $$\int\limits_{|z|=2}{2\over z^2+1}\,dz=\int\limits_{|z|=r}{2\over z^2+1}\,dz,\quad r>1$$ Next $$\left | \int\limits_{|z|=r}{2\over z^2+1}\,dz\right | \le {4\pi r\over r^2-1}\underset{r\to \infty}{\longrightarrow} 0$$