$$\oint_\Gamma \left(\frac{1}{z^2 - 2} + \frac{z}{z-5}\right) dz, \quad \text{where } \Gamma : |z+i| = \frac{1}{25}.$$
My approach
Locate singularities, namely $\sqrt{2}, -\sqrt{2}$ and $5$. As none of these fall within the circle in question is this enough to say that the integral is $0$ by Cauchy Goursat's theorem? Is this the correct approach to such a question?
Yes, your argument and your approach are correct.