calculus and analysis - Integral depends on choice of symbols

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]))

Answer

$Version

(*  "11.0.0 for Mac OS X x86 (64-bit) (July 28, 2016)"  *)

expr = (x^2 - q^2)/(z - x + I η);

i1 = Integrate [expr, 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]  *)

Change of z to w

i2 = Integrate [expr /. z -> w, 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]))  *)

Indefinite integrals can differ by an arbitrary (complex) constant of integration. The canonical order of the variables involved can affect the choice of this arbitrary constant. However, differentiation of either result returns the original expression.

expr == D[i1, x] == (D[i2, x] /. w -> z) // Simplify

(*  True  *)