Question

The value of the integral

$\int_{- 1}^{3}{\left( \tan^{- 1}\frac{x}{x^{2} + 1} + \tan^{- 1}\frac{x^{2} + 1}{x} \right)dx}$ is equal to

• A. π
• B. 2π
• C. 4π
• D. None of these

Answer 1

Answer: (B)

2π

Answer 2

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

=$\int_{- 1}^{3}{\left( \tan^{- 1}\frac{x}{x^{2} + 1} + \cot^{- 1}\frac{x}{x^{2} + 1} \right)dx}$ = $\int_{- 1}^{3}{\frac{\pi}{2}dx}$ = $2\pi$