The value of $\int_{- 1}^{3}{\left\lbrack an^{- 1}{\left( \frac{x}{x^{2} + 1} ight) + an^{- 1}\left( \frac{x^{2} + 1}{x} ight)} ight brack\ dx}$ is

Question

The value of $\int_{- 1}^{3}{\left\lbrack 	an^{- 1}{\left( \frac{x}{x^{2} + 1} ight) + 	an^{- 1}\left( \frac{x^{2} + 1}{x} ight)} ightbrack\ dx}$ is

Dr. Abubacker · Accepted Answer

(a)
$\int_{- 1}^{3}{\left\lbrack 	an^{- 1}{\left( \frac{x}{x^{2} + 1} ight) + 	an^{- 1}\left( \frac{x^{2} + 1}{x} ight)\ } ightbrack dx}$
$= \int_{- 1}^{3}{\left\lbrack 	an^{- 1}{\left( \frac{x}{x^{2} + 1} ight) + \cot^{- 1}\left( \frac{x}{x^{2} + 1} ight)} ightbrack dx}\ \ \left( \because	an^{- 1}{= \cot^{- 1}{\frac{1}{x}\ \ }} ight)$
$= \int_{- 1}^{3}{\left( \frac{\pi}{2} ight)dx = \left\lbrack \frac{\pi x}{2} ightbrack_{- 1}^{3} = \frac{\pi}{2}\lbrack 3 + 1brack = 2\pi\ \ \ \left( \because{	an^{- 1}x}{+ {\cot^{- 1}\ x =}{\frac{\pi}{2}\ \ }} ight)}$

Q. The value of $\int_{- 1}^{3}{\left\lbrack \tan^{- 1}{\left( \frac{x}{x^{2} + 1} \right) + \tan^{- 1}\left( \frac{x^{2} + 1}{x} \right)} \right\rbrack\ dx}$ is

Explanation

Reviewed by Dr. Abubacker

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