Evaluate the integral of the given contour I'm being asked to evaluate $\int \frac{1}{z^3(z^2+1)}dz$, where C is the circle $\lvert z-1 \rvert=\frac32$
I started by determining the zeroes, which are $0, -i, \,i$ Then I applied the Cauchy Integral Formula but it is here where Im not sure if i am applying it correctly. 
I took $\int \frac{1}{z^3(z^2+1)}dz$
$=\int \frac{1}{z^3(z^2+1)}dz$+$\int \frac{1}{z^3(z^2+1)}dz$+$\int \frac{1}{z^3(z^2+1)}dz$
$=2\pi i+2\pi i+2\pi i$
$=6\pi i$
Is this application correct? If not, could someone please explain to me how to do this properly
 A: Since the integrand has poles in the interior of $C$, you could  use the residue theorem or Cauchy's integral theorem instead. I think using the residue theorem would be easier.
If, however, you insist on using Cauchy's formula, write
$$\int_C \frac{dz}{z^3 (z^2 + 1)} = \int_C \frac{dz}{z^3 (z - i)(z + i)}.$$
By the Cauchy-Goursat theorem,
$$\int_C \frac{dz}{z^3 (z^2 + 1)} = \int_{C_1} \frac{(z - i)^{-1} (z + i)^{-1}}{z^3} dz + \int_{C_2} \frac{z^{-3} (z + i)^{-1}}{z - i} dz + \int_{C_3} \frac{z^{-3} (z - i)^{-1}}{z + i} dz,$$
where $C_1$, $C_2$, $C_3$, with $C_i \cap C_j = \emptyset$ for all $i \ne j$, are small circular contours around $0$, $i$, $-i$, respectively. Now, the last two integrals can be evaluated directly using Cuachy's formula. The first one, however, cannot, so you need to use Cuachy's generalized integral formula.
Alternatively, since
$$\frac{1}{z^3 (z - i)(z + i)} = \frac{1}{z^3} + \frac{1}{2(z - i)} + \frac{1}{2(z + i)} - \frac{1}{z} \tag{$\ast$}$$
and $0$, $i$, $-i$ are inside $C$,
$$\int_C \frac{dz}{2(z - i)} + \int_C \frac{dz}{2(z + i)} - \int_C \frac{dz}{z}$$
can be evaluated using Cuachy's formula. Yet, $\int_{C} z^{-3} dz$ requires Cuachy's generalized integral formula. 
