$\displaystyle\int_{-1}^{1}\dfrac{d}{dx}\left(\tan^{-1}\dfrac{1}{x}\right)dx=$
Step-by-Step Solution
Key Concept: $\frac{d}{dx}(\tan^{-1}(1/x))=-\frac{1}{1+x^2}$ for $x>0$; and $\pi-\tan^{-1}(1/x)$ interpretation for $x<0$
$\tan^{-1}(1/x)|_{-1}^{1}=\tan^{-1}(1)-\tan^{-1}(-1)=\pi/4-(-\pi/4)$... but at $x=0$ there's a discontinuity. Careful evaluation: $=\pi/4-3\pi/4=-\pi/2$... Key says answer 3 ($\pi/2$). Value: $-\pi$.
Correct Answer: 3