More and more integration

Set $a=tg(\alpha )$ Then $\frac{1}{a}=cotg(\alpha )=tg(\frac{\pi }{2}-\alpha )$ , if a>0 Or $\frac{1}{a}=cotg(\alpha )=tg(\alpha-\frac{\pi }{2} )$ , if a<0

Therefore, ${{\tan }^{-1}}(a)+{{\tan }^{-1}}(\frac{1}{a})=\frac{\pi }{2}\times sign(a)$ . The integral becomes the sum of the integral for x negative and x positive: $\int\limits_{-1}^{0}{-\frac{\pi }{2}dx}+\int\limits_{0}^{3}{\frac{\pi }{2}dx}=\pi$

$\displaystyle \int ^ { 3 } _ { -1 } ( \tan ^ { -1 } ( \frac { x } { x ^ { 2 } + 1 } ) + \tan ^ { -1 } ( \frac { x ^ { 2 } + 1 } { x } ) ) dx = \pi \rightarrow m = \boxed { 1 } .$