Question

The value of integral $\stackrel{3}{\underset{-1}{\int }}({\tan }^{-1}\left(\frac{x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{x^2+1}{x}\right))dx$

• A. $\frac{3\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. Can't be found

Answer 1

The correct option is C $2\pi$

${\int }_{-1}^3({\tan }^{-1}\left(\frac{x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{x}{1+x^2}\right))dx$

$={\int }_{-1}^3({\tan }^{-1}\left(\frac{x}{1+x^2}\right)+\cot \left(\frac{x}{1+x^2}\right))dx[\because {\tan }^{-1}\left(\frac{1}{x}\right)={\cot }^{-1}x]$

$={\int }_{-1}^3\frac{\pi }{2}dx(\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2})$

$={\int }_{-1}^3\frac{\pi }{2}dx=\frac{\pi }{2}{\left|x\right|}_{-1}^3=\frac{\pi }{2}(3-(-1\left)\right)=\frac{\pi }{2}.4=2\pi$