Question

The value of $\underset{-1}{\overset{3}{\int }}({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x})dx$ is equal to

• A. $\pi$
• B. $2\pi$
• C. $4\pi$
• D. $3\pi$

Answer 1

The correct option is A $\pi$

$\underset{-1}{\overset{3}{\int }}({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x})dx$
$=\underset{-1}{\overset{0}{\int }}({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x})dx+{\int }_0^3({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x})dx$
$=\underset{-1}{\overset{0}{\int }}-\frac{\pi }{2}dx+\underset{0}{\overset{3}{\int }}\frac{\pi }{2}dx$
$={[-\frac{\pi }{2}x]}_{-1}^0+{\left[\frac{\pi }{2}x\right]}_0^3$
$=\pi$