integrating something from a partial derivative $v=\int \frac{2x}{x^2+y^2}\,dy$ i am trying to learn harmonic analysis, and i have$$\frac{\partial u}{\partial x}=\frac{2x}{x^2+y^2}=\frac{\partial v}{\partial y}$$
and i want to get $v$. so what i do is:
$$v=\int \frac{2x}{x^2+y^2}\,dy$$
$$=\int\frac{2x}{2x}\frac{1}{a}\,da$$
$$=\ln(a)+f(x)=\ln(x^2+y^2)+f(x)$$
is that right? my problem is to do with the constant(which is just some function of $y$ since it is a partial derivative with respect to y)
should it be $\ln(x^2+y^2+f(x))$? or my above $\ln(x^2+y^2)+f(x)$?

this is from $\ln(x^2+y^2)$ but the harmonic conjugate i got was $\ln(x^2+y^2)$ which is weird! so my function is analytic at $f=\ln(x^2+y^2)+i\ln(x^2+y^2)$??

Remember that you can treat $x$ as a constant because we are integrating with respect to $y$. This motivates the following:
$$\begin{aligned}
v &= \int \frac{2x}{x^{2} + y^{2}} dy \\
&= \int \frac{2}{x \left( 1 + \left(\frac{y}{x} \right)^{2}\right)} dy \\
&= \frac{2}{x} \int \frac{dy}{\left( 1 + \left( \frac{y}{x} \right)^{2} \right)}.
\end{aligned}
$$
That last integral shouldn't be too hard, but let me know if you need more help. 
As for the constant that is troubling you, you should get $v = f(x,y) + g(x)$, where $f$ is the antiderivative above, because then we have
$$\frac{\partial v}{\partial y} = \frac{\partial f(x,y)}{\partial y} + \frac{\partial g(x)}{\partial y}.$$
This gives us $\frac{\partial f(x,y)}{\partial y} = \frac{2x}{x^{2} + y^{2}}$ and $\frac{\partial g(x)}{\partial y} = 0$; that last equality holds because $g$ is not a function of $y$.
