The value of the integral ∫−13 tan−1⁡xx2+1+tan−1⁡x2+1xdx is equal to

The value of the integral $\int_{- 1}^{3} (\tan^{- 1} \frac{x}{x^{2} + 1} + \tan^{- 1} \frac{x^{2} + 1}{x}) d x$ is equal to

A
$π$
B
$2 π$
C
$4 π$
D
none of these

Solution:

We have,

$\int_{- 1}^{3} (\tan^{- 1} \frac{x}{x^{2} + 1} + \tan^{- 1} \frac{x^{2} + 1}{x}) d x$

$= \int_{- 1}^{3} (\tan^{- 1} \frac{x}{x^{2} + 1} + \cot^{- 1} \frac{x}{x^{2} + 1}) d x = \int_{- 1}^{3} \frac{π}{2} d x = 2 π$

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