Evaluate $\displaystyle\int_0^\infty\frac{dx}{x^2+1}$
I have that $z_0=i$ and $z_1=-i$ are singularity points but just $z_0=i$ is in the upper plane then $$\int_{-\infty}^\infty\frac{dx}{x^2+1}+\int_{C_R}\frac{dz}{z^2+1}=2\pi iRes(f;z_0)$$
$\displaystyle Res(f;z_0)=\lim_{z\rightarrow i}(z-i)\frac{1}{(z-i)(z+i)}=\frac{1}{2i}$ so far so good, but now as I show that $\displaystyle\int_{C_R}\frac{dz}{z^2+1}=0?$
What you have isn't quite right.
The residue theorem gives you: $\displaystyle\int_{-R}^{R}\dfrac{dx}{x^2+1}+\int_{C_R}\dfrac{dz}{z^2+1} = 2\pi i\text{Res}(f;z_0)$. The bounds on the first integral are $-R$ to $R$ not $-\infty$ to $\infty$, since the residue theorem requires the curve to be closed.
You did correctly calculate $\text{Res}(f;z_0) = \dfrac{1}{2i}$, so $\displaystyle\int_{-R}^{R}\dfrac{dx}{x^2+1}+\int_{C_R}\dfrac{dz}{z^2+1} = \pi$ for all $R > 1$.
Taking the limit as $R \to \infty$ gives, $\displaystyle\int_{-\infty}^{\infty}\dfrac{dx}{x^2+1}= \pi - \displaystyle\lim_{R \to \infty}\int_{C_R}\dfrac{dz}{z^2+1}$ for all $R > 1$.
So, what you need to show is that $\displaystyle\lim_{R \to \infty}\int_{C_R}\dfrac{dz}{z^2+1} = 0$.
To do that, notice that $0 \le \displaystyle\left|\int_{C_R}\dfrac{dz}{z^2+1}\right| \le \int_{C_R}\dfrac{|dz|}{|z^2+1|} \le \int_{C_R}\dfrac{|dz|}{|z^2|-1} = \int_{C_R}\dfrac{|dz|}{|z|^2-1} = \int_{C_R}\dfrac{|dz|}{R^2-1}$ $= \displaystyle\dfrac{1}{R^2-1}\int_{C_R}|dz| = \dfrac{1}{R^2-1}\text{length}(C_R) = \dfrac{\pi R}{R^2-1}$.
Do you see how this helps?