For the given u(x,y) find a v(x,y) such that f(z) = u(x,y) + iv(x,y) is holomorphic in some region of $\mathbb{C}$ I used the Cauchy Riemann equations and got two different functions v(x,y) by integrating.  Should I just add them up?  
My u(x,y) is $\frac{x}{x^2 + y^2}$
My partial wrt x is $\frac{y^2 - x^2}{x^2 + y^2}$ and my partial wrt y is $\frac{-2xy}{x^2 + y^2}$.
 A: $u(x,y)$ partial wrt $x$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $y$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
By the rule of Cauchy-Riemann Equations,we have
$v(x,y)$ partial wrt $y$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $x$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
Then we can let $v(x,y)=\frac{-y}{x^2+y^2}$.
A: First, your partial derivatives are a bit off. You should have:
$$\frac{\partial u}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2}, \quad \frac{\partial u}{\partial y} = \frac{-2xy}{(x^2+y^2)^2}$$
Next, remember what the Cauchy-Riemann Equations say:
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$
So we see that the Cauchy-Riemann Equations dictate that $v$ must satisfy 
$$\frac{\partial v}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2},\quad \frac{\partial v}{\partial x} = \frac{2xy}{(x^2+y^2)^2}$$
Try constructing $v$ from this.
A: While you already got an answer for your question, i will give a detailed solution for the sake of completeness.
But first let me point out that your aim is to find the harmonic conjugate of the function $u$ you mention (check that it is indeed harmonic). For the one familiar withe DEs, all of this is just routine.
Now for the construction of $v$:
$$\frac{\partial u}{\partial x}=\frac{y^{2}-x^{2}}{(x^2+y^2)^2}\quad and\quad\frac{\partial u}{\partial y}=-\frac{2yx}{(x^2+y^2)^2}$$
The Cauchy-Riemann equations must be satisfied, i.e.
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad and\quad\frac{\partial u}{\partial x}=-\frac{\partial u}{\partial y}$$
Thus:
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\Leftrightarrow\frac{y^{2}-x^{2}}{(x^2+y^2)^2}=\frac{\partial v}{\partial y}$$
Integrating both sides of the above equation (wrt $y$) we get
$$v(x,y)=-\frac{y}{x^2+y^2}+c(x)\quad(1)$$
where $c$ is a (smooth) function of $x$.
Next, differentiate both sides of $(1)$ wrt $x:$
$$\frac{\partial v}{\partial x}=\frac{2xy}{(x^2+y^2)^2}+c'(x)\quad(2)$$
Use $(2)$ and the right C-R equation to obtain $c'(x)=0,$ thus $c$ is a constant function. Then $(1)$ gives the desired $v:$
$$v(x,y)=-\frac{y}{x^2+y^2}+c$$
