The value of $\int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x$, is

$-\pi / 2$

We have,

$\frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right)=\frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{1}{1+x^2}$

∴   $\int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x=\int\limits_{-1}^1-\frac{1}{1+x^2} d x=-2 \int\limits_0^1 \frac{1}{1+x^2} d x$

$\Rightarrow \int\limits_{-1}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x=-2\left[\tan ^{-1} x\right]_0^1=-2\left(\frac{\pi}{4}\right)=-\frac{\pi}{2}$