Question

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

Answer 1

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

Now, ${\tan }^{-1}\frac{x}{x^2+1}+{\tan }^{-1}\frac{x^2+1}{x}={\tan }^{-1}\frac{\frac{x}{x^2+1}+\frac{x^2+1}{x}}{1-(\frac{x}{x^2+1}\times \frac{x^2+1}{x})}$

$={\tan }^{-1}\left(\infty \right)=\frac{\pi }{2}$.

$\therefore$ From $\left(1\right),I={\int }_{-1}^3\frac{\pi }{2}dx$.

$=\frac{\pi }{2}{\left[x\right]}_{-1}^3=\frac{\pi }{2}[3+1]=\frac{\pi }{2}\times 4=2\pi$ Ans