<s>[INST] Evaluation of $\int_\limits{-1}^{3}\frac{x\ln |x|}{1+x^4}dx$ 
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
 [/INST] 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}
</s>