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The original integral clearly diverges logarithmically "at $\ds{x = 0}$". We'll evaluate the Principal Value by assuming it was the OP original intention.
\begin{align}
&\mrm{P.V.}\int_{-1}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{-1}^{-\epsilon}{\arctan\pars{1 + x^{2}} \over x}\,\dd x +
\int_{\epsilon}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x}
\\[5mm] = &\ \
\overbrace{\mrm{P.V.}\int_{-1}^{1}{\arctan\pars{1 + x^{2}} \over x}\,\dd x}
^{\ds{=\ 0}}\ +\
\int_{1}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x\quad
\pars{\begin{array}{c}
\mbox{Integrate by parts}\\ \mbox{the last integral}
\end{array}}
\\[5mm] = &\
\ln\pars{3}\arctan\pars{10} -
\int_{1}^{3}\ln\pars{x}\,{2x \over \pars{1 + x^{2}}^{2} + 1}\,\dd x
\\[5mm] \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=} &\,\,\,
\ln\pars{3}\arctan\pars{10} -
{1 \over 2}\int_{1}^{9}{\ln\pars{x} \over \pars{1 + x}^{2} + 1}\,\dd x
\\[5mm] = &\
\ln\pars{3}\arctan\pars{10} -
{1 \over 2}\int_{1}^{9}\ln\pars{x}
\pars{{1 \over x + 1 - \ic} - {1 \over x + 1 + \ic}}{1 \over 2\ic}\,\dd x
\\[5mm] = &\
\ln\pars{3}\arctan\pars{10} +
{1 \over 2}\,\Im\int_{1}^{9}{\ln\pars{x} \over -1 + \ic - x}\,\dd x
\\[5mm] = &\
\ln\pars{3}\arctan\pars{10} +
{1 \over 2}\,\Im\int_{1}^{9}{%
\ln\pars{\bracks{-1 + \ic}\braces{x/\bracks{-1 + \ic}}} \over
1 - x/\pars{-1 + \ic}}\,{\dd x \over -1 + \ic}
\\[5mm] \stackrel{x/\pars{-1 + \ic}\ \mapsto\ x}{=} &\,\,\,
\ln\pars{3}\arctan\pars{10} +
{1 \over 2}\,\Im\int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2}{%
\ln\pars{\bracks{-1 + \ic}x} \over 1 - x}\,\dd x
\\[1cm] = &\
\ln\pars{3}\arctan\pars{10} +
{1 \over 2}\,\Im\left\lbrace%
-\ln\pars{1 + {9 \over 2}\,\bracks{1 + \ic}}\ln\pars{9}\right.
\\[5mm] &\phantom{\ln\pars{3}\arctan\pars{10} +
{1 \over 2}\,\Im\braces{}} +
\left.\int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2}{%
\ln\pars{1 - x} \over x}\,\dd x\right\rbrace
\\[1cm] = &\
\ln\pars{3}\arctan\pars{10} - \ln\pars{3}
\arctan\pars{9 \over 11} -
{1 \over 2}\,\Im
\int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2}
\mrm{Li}_{2}'\pars{x}\,\dd x
\\[5mm] = &\
\bbox[#ffe,15px,border:1px dotted navy]{\ds{{1 \over 4}\,\ln\pars{3}\,\pi -
{1 \over 2}\,\Im\braces{%
\mrm{Li}_{2}\pars{-\,{9 \over 2}\,\bracks{1 + \ic}} -
\mrm{Li}_{2}\pars{-\,{1 + \ic \over 2}}}}}\ \approx 1.4421
\end{align}
Note that
$\ds{\arctan\pars{10} - \arctan\pars{9/11} =
\arctan\pars{\bracks{10 - 9/11}/\braces{1 + 10\bracks{9/11}}} =
\arctan\pars{1} = \pi/4}$.