Find the real part of the following analytic function Given $v = \frac{-y}{x^2+y^2}$, find the real component of the function $u$. 
I'm not sure why I'm not getting this. Here's my work so far:
$$v_x = \frac{2xy}{(x^2+y^2)^2} = -u_y \Rightarrow u(x,y) = \frac{x}{x^2+y^2} + \phi(x) $$
where $\phi(x)$ is (as of yet) undetermined. Continuing on, I get
$$ u_x = \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} + \phi'(x) = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} = v_y $$
simplifying
$$ \phi'(x) = \frac{2x^2+2y^2}{(x^2+y^2)^2} - \frac{2}{x^2+y^2}$$
multiplying the rightmost fraction by $x^2+y^2$ (top and bottom) we find that
$$\phi'(x) = 0 \Rightarrow \phi(x) = C \in \mathbb{R}$$
so
$$ u(x,y) = \frac{x}{x^2+y^2} + C$$
This satisfies first first of the Cauchy-Reimann conditions, but it does not seem to satisfy the second. Explicitly the condition that $u_x = v_y$ is
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} $$
I don't believe that this is generally true… The only idea that I have is that perhaps $\phi(x) = C$ is a constant with respect to $x$, but not $y$ but I think that would contradict my first line of reasoning above. 
Any ideas?

Hint:
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2}  \iff$$
$$\iff  \frac{2}{x^2+y^2} = \frac{2(x^2+y^2)}{(x^2+y^2)^2} $$
So your work seems correct !
