Question

${\int }_{-1}^1\frac{d}{dx}\left(ta{n}^{-1}\frac{1}{x}\right)dx$ is

• A. $\frac{\pi }{2}$
  
• B. $\frac{\pi }{4}$
  
• C. $-\frac{\pi }{2}$
  
• D. $-\frac{\pi }{4}$

Answer 1

The correct option is C $-\frac{\pi }{2}$


${\int }_{-1}^1\frac{d}{dx}\left(ta{n}^{-1}\frac{1}{x}\right)dx$
${\int }_{-1}^1\frac{1}{1+{\left(\frac{1}{x}\right)}^2}\left(\frac{-1}{x^2}\right)dx=-{\int }_{-1}^1\frac{1}{1+x^2}dx=-2{\int }_0^1\frac{dx}{1+x^2}$ (even function)
$=-2{\left[ta{n}^{-1}x\right]}_0^1=-2\frac{\pi }{4}=-\frac{\pi }{2}$