Evaluation of $$\int_\limits{-1}^{3}\frac{\arctan(1+x^2)}{x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_\limits{-1}^3\arctan(1+x^2)\cdot \frac{1}{x}dx$$
Using By parts, We get
$$I = \left[\arctan (1+x^2)\cdot \ln|x|\right]^{3}_{-1}-2\int_\limits{-1}^3\frac{x\ln |x|}{1+x^4}dx$$
Now How can i solve it , Help required, Thanks
By continuity, there exists a neighborhood about zero such that \begin{align} |\arctan 1 -\arctan (1+x^2)| < \frac{1}{4} \ \ \Rightarrow \ \ \arctan1<\frac{1}{4}+\arctan (1+x^2) \end{align} which means \begin{align} 0<\frac{\frac{\pi}{4}-\frac{1}{4}}{x} < \frac{\arctan(1+x^2)}{x} \end{align} whenever $r<x<0$ for some $r$. Now, observe \begin{align} \infty =\int^0_{-1} \frac{\frac{\pi-1}{4}}{x}\ dx< \int^0_{-1} \frac{\arctan(1+x^2)}{x}\ dx. \end{align}
$$\arctan(x^2+1) > \arctan (1)=\frac{\pi}{4}$$
For $x \in (0,\infty)$ as $x^2 > 0 \implies x^2+1 > 1$ and $f(u)=\arctan(u)$ is strictly increasing on $x \in (0,\infty)$.
Hence,
$$\int_{0}^{3} \frac{\arctan (x^2+1)}{x} > \int_{0}^{3} \frac{\pi}{4} \frac{1}{x} \to \infty$$
The integral does not converge as has been explained above, however it does have a finite principal value (apparently PV is $\frac{31423231}{21789317}$ or 1.44214 )
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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
$\ds{\arctan\pars{10} - \arctan\pars{9/11} = \arctan\pars{\bracks{10 - 9/11}/\braces{1 + 10\bracks{9/11}}} = \arctan\pars{1} = \pi/4}$.
