I am trying to perform integrals of the kind $\int dx\frac{x^2-q^2}{z-x+i\eta}$. Mathematica, however, gives back different results whenever I replace the parameter $z$ by $w$, for instance, as follows
Integrate[$\frac{x^2-q^2}{z-x+i \eta}$,$x$]//FullSimplify=$-\frac{1}{2}(x-z-i\eta)(x+3z+3i\eta)+(q^2-(z+i\eta)^2)\log{\left(-x+z+i\eta\right)}$ and
Integrate[$\frac{x^2-q^2}{w-x+i \eta}$,$x$]//FullSimplify=$\frac{1}{2}\left(-x\left(2w+x+2i\eta\right)+\left(q-w-i\eta\right)\left(q+w+i\eta\right)\left(2i\arctan{\left(\frac{\eta}{w-x}\right)}+\log{\left(\left(w-x\right)^2+\eta^2\right)}\right)\right)$
The results do not only have a different form, but taking the limit $\eta\rightarrow0$ after performing the integral yields different results. Note that the only difference between the integrals is the name of the parameter $z$. Can anyone tell me what's going on here and what the correct result would be. Thanks!
(Code)
Integrate[(x^2 - q^2)/(z - x + I η), x] // FullSimplify
-(1/2) (x - z - I η) (x + 3 z + 3 I η) + (q^2 - (z + I η)^2) Log[-x + z + I η]
Integrate[(x^2 - q^2)/(w - x + I η), x] // FullSimplify
1/2 (-x (2 w + x + 2 I η) + (q - w - I η) (q + w + I η) (2 I ArcTan[η/(w - x)] + Log[(w - x)^2 + η^2]))