I have the following problem which I want to evaluate at infinity: $$\oint \dfrac{(z+2)}{(z^2+9)}dz$$
I approach this problem by saying that $z=\dfrac{1}{t}$ and $dz=\dfrac{-1}{t^2}dt$. And I plug them inside my integral and obtain: $$\oint \dfrac{-(2t+1)}{(t+9t^3)}dz=-2\pi i Res(0)=-2\pi i$$ Yet this result is not in accordance with the usual integration using residues which yield $2\pi i$. I was wondering where am I doing a mistake of minus.
On
If you mean you want $\displaystyle \oint_\Gamma \dfrac{z+2}{z^2+9} \; dz$ where $\Gamma$ is a positively oriented contour near $\infty$ (i.e. outside the poles at $z = \pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $\infty$:
$$ \dfrac{z+2}{z^2+9} = \dfrac{z+2}{z^2} (1 - 9/z^2 + \ldots) = \frac{1}{z} -\frac{7}{z^2} + \ldots $$
so the result is $2\pi i$.
If you are trying to evaluate the integral,$\oint_{|z|=R}\frac{z+2}{z^2+9}\,dz$, after making the transformation $z\mapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $z\mapsto 1/z$
$$\begin{align} \oint_{|z|=R}\frac{z+2}{z^2+9}\,dz&=\oint_{|z|=1/R}\frac{1+2z}{z(1+9z^2)}\,dz\\\\ &=2\pi i \text{Res}\left(\frac{1+2z}{z(1+9z^2)},z=0 \right)\\\\ &=2\pi i \end{align}$$
as expected.
Recall that under the transformation, $dz\mapsto -\frac1{z^2}\,dz$ and the contour is traversed clockwise.