Let's do something wacky and fun
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Note that $\displaystyle tan^{-1}(u)=\frac{i}{2}\left(ln(1-ix)-ln(1+ix)\right)$
$\displaystyle =\frac{i}{2}ln(\frac{1-ix}{1+ix})$
$\displaystyle =\frac{i}{2}ln\left(\lim_{x\to 0}\frac{2-2i\cdot ln(x)}{1+(ln(x))^{2}}-1\right)$
Use L'hopital on the inside:
$\displaystyle \frac{i}{2}ln\left(\lim_{x\to 0}\frac{-i}{ln(x)}-1\right)$
The limit tends to -1 and we get $\displaystyle \frac{i}{2}ln(-1)$
$\displaystyle ln(-1)={\pi}i$
So, we get $\displaystyle \frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}$