I have an integral of the following form $$ \int^w_0 \frac{bz}{(z-a)(z+a)}\textrm{d}z,\quad z\in\mathbb{C} $$ which goes from the origin to the point $w$ in the complex-plane. I'm feeling a bit shaky about how to solve this integral. I'm thinking about parameterise $z$, but I don't really know if that will help me. I'm also aware of the two poles that I have at $z=\pm a$.
A clue of how to get started would be of great appreciation.
Background: I have a scalar potential given by $\phi$ that's a function of two complex variables $\alpha$ and $\beta$. Such a potential defines a generalised force $$\vec{\Phi}(\alpha,\beta)=-\vec{\nabla}\phi(\alpha,\beta)$$ For the function $\phi$ to be a well-behaved potential one must require that the crossed derivatives of the force components are the same $$ \frac{\partial}{\partial\alpha}\Phi_\beta= \frac{\partial}{\partial\beta}\Phi_\alpha $$ In my case the force vector is given by $$ \vec{\Phi}(\alpha,\beta)= 2 \begin{pmatrix} \dfrac{k\beta\alpha^2+(i\kappa/4-k)\alpha-\beta\varepsilon_p}{k\alpha^2-\varepsilon_p}\\ \dfrac{-k\alpha\beta^2+(i\kappa/4+k)\beta+\alpha\varepsilon^*_p}{-k\beta^2-\varepsilon^*_p} \end{pmatrix} $$ and the scalar potential can be found from $$ \phi(\alpha,\beta)=\phi(0,0)-\underbrace{\int^{(\alpha,0)}_{(0,0)}\Phi_\alpha(\alpha',0)d\alpha'}_{\textrm{I}} -\int^{(\alpha,\beta)}_{(\alpha,0)}\Phi_\alpha(\alpha,\beta')d\beta' $$ Focusing on the integral I we have $$ \int^{(\alpha,0)}_{(0,0)}\Phi_\alpha(\alpha',0)d\alpha' =2\int^{(\alpha,0)}_{(0,0)}\frac{(i\kappa/4-k)\alpha'}{k\alpha^{\prime 2}-\varepsilon_p}d\alpha' $$ which leads back to my original problem if we call $b=(i\kappa/4-k)$, $a=\sqrt{\varepsilon/k}$ and $z=\alpha'$
Hint:
$$\frac1{z-a}+\frac1{z+a}=\frac{2z}{(z-a)(z+a)}$$ leads to the antiderivative
$$\log(z-a)+\log(z+a)=\log(x^2-y^2-a^2+i2xy).$$
Now you have to discuss the handling of the singularities.