Find $\lim_{a\to \infty}\frac{1}{a}\int_0^{\infty}\frac{x^2+ax+1}{1+x^4}\cdot\arctan(\frac{1}{x})dx$ Find
$$
\lim_{a\to \infty}
    \frac{1}{a}
    \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(\frac{1}{x}\right)dx
$$

I tried to find 
$$
\int_0^{\infty} \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx
$$
Let
$$
I(a) = \int_0^\infty
           \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx
$$
Let
$$
\begin{split}
t &= \arctan(1/x) \\
\frac{dt}{dx} &= \frac{-1}{1+x^2} \\
dt &= \frac{-1}{1+x^2}dx
\end{split}
$$
I am stuck here, there seems no way to further solving.
 A: One may observe that, as $a\to \infty$,
$$
\begin{align}
&\frac1a
    \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx-
   \color{red}{\int_0^{\infty}\!\!\frac{x}{1+x^4} \arctan\left(1/x\right)\:dx}
\\\\&=\frac1a\int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx-
    \frac1a\int_0^{\infty}\frac{ax}{1+x^4} \arctan\left(1/x\right)\:dx
\\\\&=\frac1a\int_0^{\infty}\frac{x^2+1}{1+x^4} \arctan\left(1/x\right)\:dx
\: \longrightarrow \: 0,\tag1
\end{align}
$$ since the latter integral is convergent. 
On the other hand, we have
$$
\int_0^{\infty}\frac{x}{1+x^4} \:\arctan\left(1/x\right)\:dx=\int_0^{\infty}\frac{x}{1+x^4}\: \arctan\left(x\right)\:dx\quad (x \to 1/x)
$$ using, $\displaystyle \arctan\left(x\right)+\arctan\left(1/x\right)=\frac{\pi}2$ ($x>0$), one gets
$$
\begin{align}
&\int_0^{\infty}\!\!\frac{x}{1+x^4}\: \arctan\left(x\right) \:dx+\int_0^{\infty}\!\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx
\\\\& =\frac{\pi}2\int_0^{\infty}\!\!\frac{x\:dx}{1+x^4}
\\\\&=\frac{\pi}4\int_0^{\infty}\!\!\frac{du}{1+u^2}
\\\\& =\frac{\pi^2}8,
\end{align}
$$ thus 
$$\displaystyle  \color{red}{\int_0^{\infty}\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx}=\frac{\pi^2}{16}$$ giving, from $(1)$,

$$
\lim_{a\to \infty}
    \frac{1}{a}
    \int_0^{\infty}\frac{x^2+ax+1}{1+x^4}\: \arctan\left(\frac{1}{x}\right)dx=\frac{\pi^2}{16}.\tag2
$$

