The integral $\int_{- 1}^{3}\left\lbrack an^{- 1}{\frac{x}{x^{2} + 1} + an^{- 1}\frac{x^{2} + 1}{x}} ight brack\ dx$ is equal to

Question

The integral $\int_{- 1}^{3}\left\lbrack 	an^{- 1}{\frac{x}{x^{2} + 1} + 	an^{- 1}\frac{x^{2} + 1}{x}} ightbrack\ dx$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Let $I = \int_{- 1}^{3}\left\lbrack 	an^{- 1}{\frac{x}{x^{2} + 1} + 	an^{- 1}\frac{x^{2} + 1}{x}} ightbrack\ dx$
Now, $	an^{- 1}\frac{x}{x^{2} + 1} + 	an^{- 1}\frac{x^{2} + 1}{x}$
$= 	an^{- 1}\left\lbrack \frac{\frac{x}{x^{2} + 1} + \frac{x^{2} + 1}{x}}{1 - \frac{x}{x^{2} + 1}.\ \frac{x^{2} + 1}{x}} ightbrack$
$= 	an^{- 1}\left\lbrack \frac{x^{2} + \left( x^{2} + 1 ight)^{2}}{x\left( x^{2} + 1 ight).0} ightbrack = 	an^{- 1}(\infty) = \frac{\pi}{2}$
Hence, $I = \int_{- 1}^{3}\frac{\pi}{2}\ dx = 2\pi$

Q. The integral $\int_{- 1}^{3}\left\lbrack \tan^{- 1}{\frac{x}{x^{2} + 1} + \tan^{- 1}\frac{x^{2} + 1}{x}} \right\rbrack\ dx$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?