Let $x=ae^y$ then we have to show that

$\tan^{-1} i\left ( \dfrac {ae^y -a}{ae^y +a} \right ) = \tan^{-1} \left ( \dfrac {e^y - 1}{e^y + 1} \right ) = \dfrac {i}{2} \log e^y = \dfrac {iy}{2}$

i.e. we have to show that

$i\dfrac {e^y - 1}{e^y + 1} = \tan \dfrac {iy}{2} = i \tanh \dfrac {y}{2}$

$\text{But }\tanh \dfrac {y}{2} = \dfrac {e^{\frac {y}{2}} - e^{-\frac {y}{2}}}{e^{\frac {y}{2}} + e^{-\frac {y}{2}}} = \dfrac {e^y-1}{e^y+1}$

Hence the result.