$$f(z)=\frac{1}{z+1}+\frac{1}{z-1}$$
$$\gamma_j:[0,2\pi]\rightarrow \mathbb{C} \ (j=1,2,3)$$
\begin{align} \gamma_j(t) & =(-1)^j+\frac{1}{2}\exp(it) \text{ for } j=1,2 \\[10pt] \gamma_3(t) & =4\exp(it) \end{align}
I need to compute
$$\int_{\gamma_1}f(z)\, \mathrm{d}z + \int_{\gamma_2}f(z) \,\mathrm{d}z \text{ and } \int_{\gamma_3}f(z) \,\mathrm{d}z.$$
Can I compute them using $$\int_\gamma{dz\over z-a}=2\pi i\> n(\gamma,a)$$
And if so, how is it done?
On
Notice that $f(z)$ has poles at $1$ and $-1$.
You should also know that for a path parametrized as $$\gamma(t)=s+re^{it}$$ is a circle of radius $r$ centered at $s$.
Also note that in the formula you stated at the end of your question, $n(\gamma,a)$ is $0$ if $\gamma$ does not contain $a$, else equal $k$ where $k$ is the number of times the path is travelled.
Then one can immediately see $\gamma_1$ contains the pole at $-1$ only and $\gamma_2$ contains the pole at $1$ only.
Thus, you get the answer $2\pi i$ for the first two integrals(conventionally the path is travelled once so $n(\gamma,a)=1$).
$\gamma_3$ contains both poles. But don’t panic.
$$\int_{\gamma_3} f(z) \, dz=\int_{\gamma_3} \frac1{z+1}\,dz+\int_{\gamma_3} \frac1{z-1} \, dz$$
Do you now see the answer?
The curve $\gamma_1$ winds once counterclockwise around $-1.$ It's just a circle of radius $1/2.$ So you can apply the equality you state at the end of the question to $\displaystyle \int_{\gamma_1} \frac{dz}{z+1}.$ But you have the other term, $\displaystyle \int_{\gamma_1} \frac{dz}{z-1}.$ Here you need to note that $z\mapsto\dfrac 1 {z-1}$ is holomorphic everywhere in the interior and on the boundary of $\gamma_1.$ There is a theorem dealing with that situation.
As for $\gamma_3,$ have you drawn the circle and observed how it is related to the two points $1$ and $-1\text{?}$