Question

Evaluate ${\int }_{-1}^3({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x})dx$.

Answer 1

$\begin{array}{c}{\int }_{-1}^3\left({\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x}\right)dx\\ Useidentity{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\\ \Rightarrow {\int }_{-1}^3{\tan }^{-1}\propto dx={\int }_{-1}^3\frac{\pi }{2}dx\\ \Rightarrow \frac{\pi }{2}{\left[x\right]}_{-1}^3=2\pi \end{array}$

Hence, this is the answer.