$\underset{-1}{\overset{3}{\int }}\left(Ta{n}^{-1}\frac{{\displaystyle x}}{{\displaystyle x^2+1}}+Ta{n}^{-1}\frac{{\displaystyle x^2+1}}{{\displaystyle x}}\right)dx=$

$$\begin{gathered} {\int }_{-1}^3\left({\tan }^{-1}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}+{\tan }^{-1}\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)\mathrm{dx} \\ ={\int }_{-1}^0\left({\tan }^{-1}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}+{\tan }^{-1}\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)+{\int }_0^3\left({\tan }^{-1}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}+{\tan }^{-1}\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right) \\ ={\int }_{-1}^0 \frac{-\mathrm{\pi }}{2}\mathrm{dx}+{\int }_0^3 \frac{\mathrm{\pi }}{2}\mathrm{dx} \\ =\left(\frac{-\mathrm{\pi }}{2}\right)(0-(-1\left)\right)+\frac{\mathrm{\pi }}{2}(3-0) \\ =\frac{-\mathrm{\pi }}{2}+\frac{3\mathrm{\pi }}{2}=\mathrm{\pi } \end{gathered}$$