Without residue theory, determine $$\oint \limits_{C} \frac{2}{z^3+z}dz$$ if $C: \big|~z~-~\frac{i}{2}~\big|=1$ is positively oriented.
We first find that our integrand has three distinct singular points $z=\{-i,~0,~i\}$
Now if we draw a sketch of $C$, we see that only two of these singular points are within $C$.
We can thus rewrite our integral as $$\oint \limits_{C} \frac{2}{z^3+z}dz = 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz$$
Now we may define two closed, piecewise smooth curves $C_1, C_2$ around each singular point in $C$ as follows
We then have that \begin{align}\oint \limits_{C} \frac{2}{z^3+z}dz &= 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz \\ &= 2 \bigg[ \underbrace{\oint \limits_{C_1} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_1} + \underbrace{\oint \limits_{C_2} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_2} \bigg] \end{align}
Now let $f(z) = \frac{1}{z(z+i)}$ then, from Cauchy Integral Formula, we know that \begin{align}I_1 &=\oint \limits_{C_1} \frac{f(z)}{(z-i)}dz \\ &= 2\pi i ~f(i) \\ &=-\pi i\end{align}
Now let $g(z) = \frac{1}{(z+i)(z-i)}$, then from Cauchy Integral Formula, we know that \begin{align}I_2 &= \oint \limits_{C_2} \frac{g(z)}{z}dz \\ &= 2\pi i ~ g(0) \\ &= 2\pi i\end{align}
So finally we have that \begin{align}\oint \limits_C \frac{2}{z^3 + z}dz = 2\big( -\pi i + 2\pi i\big) = 2 \pi i\end{align}
Does this seem correct?
It's correct. Essentially what you've done is exactly how the proof of the residue theorem goes anyway - you surround each isolated pole with a small contour and apply Cauchy's theorem to each individual pole. The result is the sum of the residues, multiplied by $2\pi i$, which is the residue theorem. Now you know why the residue formula is what it is!