Question

$\int_{- 1}^{3}{\left\lbrack \tan^{- 1}\frac{x}{x^{2} + 1} + \tan^{- 1}\frac{x^{2} + 1}{x} \right\rbrack dx}$ =

• A. π
• B. 2π
• C. 3π
• D. None of these

Answer 1

Answer: (B)

2π

Answer 2

$I = \int_{- 1}^{3}\left\lbrack \tan^{- 1}\frac{x}{x^{2} + 1} + \cot^{- 1}\frac{x}{x^{2} + 1} \right\rbrack dx$

⇒ $I = \int_{- 1}^{3}{\frac{\pi}{2}dx}$ ⇒ $I = \frac{\pi}{2}\lbrack x\rbrack_{- 1}^{3} = \frac{\pi}{2}\lbrack 3 + 1\rbrack = 2\pi$.