how to find analytic function? 
If $v=\Large \frac{x-y}{x^2+y^2}$ then the analytic function $f(z)=u(x, y)+iv(x, y)$ is
a) $z+c\qquad $  b) $z^{-1}+c\qquad $ c) $\frac{1-i}{z}+c\qquad $  d) $\frac{1+i}{z}+c\qquad $

my try: i computed
$$\frac{\partial v}{\partial x}=\frac{y^2-x^2+2xy}{(x^2+y^2)^2}$$
$$\frac{\partial v}{\partial y}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}$$
for analytic function $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ so i get
$$\frac{\partial u}{\partial x}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}$$
$$\int \partial u=\int \frac{y^2-x^2-2xy}{(x^2+y^2)^2}\partial x$$
$$u=\int\left(\frac{1}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2} -\frac{2xy}{(x^2+y^2)^2}\right)\partial x$$
i got stuck here i don't know how to find $u$ & then $f(z)$. can somebody please give some easy method to find analytic function $f(z)$, thank you very much

Let us suppose $u$ is of a similar form to $v$.
$$u = \frac{ax + by}{x^2 + y^2} + c$$
$$\frac{\partial u}{\partial x} = \frac{a(x^2 + y^2) - 2x(ax+by)}{(x^2 + y^2)^2} = \frac{ay^2 -ax^2 - 2bxy}{(x^2 + y^2)^2}$$
Equating coefficients with $\frac{\partial v}{\partial y}$ gives $a = 1,b = 1$. At this point it is worth checking the other Cauchy-Riemann equation to confirm the function is indeed analytic. It remains to put the function into the correct form:
$$f(z) = \frac{x+y}{x^2 +y^2} + i\frac{x-y}{x^2+y^2} + c= \frac{(1+i)\bar{z}}{z\bar{z}} + c = \frac{1+i}{z} + c$$
