\dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x} =-\dfrac{1}{1+x^2}
But antiderivative of \dfrac{1}{1+x^2} is \cot^{-1}x
So an alternative presentation of the solution would be
\tan^{-1} \dfrac{1}{x} = \cot^{-1}x ,x>0
= \cot^{-1}x - \pi, x<0
We see that \dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x}=\dfrac{\mathrm{d}}{\mathrm{dx}} \cot^{-1}x in either case
Hence \displaystyle \int_{-1}^1 \dfrac{\mathrm{d}}{\mathrm{dx}}\tan^{-1} \dfrac{1}{x} \ \mathrm{dx}= \int_{-1}^1 \dfrac{\mathrm{d}}{\mathrm{dx}} \cot^{-1}x\ \mathrm{dx}
= \displaystyle |\cot^{-1} x|_{-1}^1 = \dfrac{\pi}{4} - \dfrac{3\pi}{4}=-\dfrac{\pi}{2}
