So I have found out $$\frac{\partial v}{\partial x} = \frac{2x-2}{(x-1)^2 +(y-2)^2},\ \frac{\partial v}{\partial y} = \frac{2y-4}{(x-1)^2 +(y-2)^2}.$$
Using the Cauchy Riemann equations, I find: $$\frac{\partial u}{\partial x} = \frac{2y-4}{(x-1)^2 +(y-2)^2},\ \frac{\partial u}{\partial y} = \frac{2-2x}{(x-1)^2 +(y-2)^2}.$$
Then I integrate either one of $\partial u/\partial x$ or $\partial u/\partial y$ to then differentiate with respect to the other variable I just integrated with. However, when I integrate $\partial u/\partial x$ or $\partial u/\partial y$, I get two different functions: $2\arctan\frac{(x-1)}{(y-2)} + g(y)$ and $-2\arctan\frac{(y-2)}{(x-1)} + g(x)$, respectively.
What am I doing wrong? Can you have multiple harmonic conjuguates?
You will not be able to find an harmonic conjugate of $v$ since there is none. That is, there is no function $u\colon\mathbb{C}\setminus\{1+2i\}\longrightarrow\mathbb C$ which is an harmonic conjugate of $v$. You will find a proof here.