Breaking a contour integral into 3 separate contours? We can try to integrate the following function around a counter-clockwise circular contour:
$$\frac{x^3}{(x-1)(x-2)(x-3)}$$
Can someone show how to use the Cauchy–Goursat theorem (explained here and here) to break this apart into 3 separate contours?
In other words, I'd like to take
$$\int_c{\frac{x^3}{(x-1)(x-2)(x-3)}}$$
and get
$$\int_{C_1}{f_1(x)} + \int_{C_2}{f_2(x)} + \int_{C_3}{f_3(x)}$$
I'm hoping for a pretty thorough explanation with at least one of the contours.  I just want to be certain I have the idea perfected.
 A: If you're looking to break apart the contour into three parts so that each part contains exactly one pole, take a look at the Mercedes-Benz symbol. Imagine the circle is the original contour. Your function $f(z)$ has three poles, so we subdivide the circle into three parts so that each part corresponds to one of the contours, each of which contains exactly one pole.
So now we have our $C_1, C_2, C_3$ contours. So then we can break up our original integral into
$$ \int_{C_1} f(z) \, dz + \int_{C_2} f(z) \, dz + \int_{C_3} f(z) \, dz$$
So the integrand isn't actually different. However, because each contour contains only one pole, we can take advantage of the integral formula as follows: suppose we are looking at the $C_1$ contour; then $z = 1$ is the only pole, so then we should write
$$ f(z) = \frac{1}{z - 1} \left( \frac{ x^3}{(x - 2)(x - 3)} \right) = \frac{f_1(z)}{z - 1}$$
then
$$ \int_{C_1} f(z) \, dz = \int_{C_1} \frac{f_1(z)}{z - 1} \, dz = 2\pi i f_1(1) $$
according to the integral formula. We repeat this same process similarly for the other contours.
