Evaluate $\lim_{n\to\infty} \int_{n}^{\infty}\frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2+x^2}$ I need to evaluate $$\lim_{n\to\infty} \int_{n}^{\infty}\frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2+x^2}$$
I tried to apply DCT, and this is my work
$$ \lim_{n\to\infty} \int_{n}^{\infty}\frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2+x^2}dx=\lim_{n\to\infty}\int_{0}^{\infty}\frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2+x^2}\mathbb{1}_{[n,\infty)}dx$$
Then we have
$$\begin{split}
\frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2+x^2} \mathbb{1}_{[n,\infty)}&\leq \frac{n^2\arctan(x)\arctan(\frac{1}{x})}{n^2} \mathbb{1}_{[n,\infty)} \\
&\leq \arctan(x)\arctan(\frac{1}{x}) \\
&\leq\frac{{\pi}^2}{16}
\end{split}$$
But $\frac{{\pi}^2}{16}$ is not integrable over $[0,\infty)$. Did I make any mistake? Can this task be solved by this method?

One has:
\begin{align}
I_n:=\int^\infty_n \frac{n^2\arctan(x)\arctan(1/x)}{n^2+x^2}\,dx=\int^\infty_n \frac{\arctan(x)\arctan(1/x)}{1+(x/n)^2}\,dx
\end{align}
Substitute $t=x/n$ so that $n dt=dx$ and hence:
\begin{align}
I_n = \int^\infty_1 \frac{n\arctan(nt)\arctan\left(\frac{1}{nt}\right)}{1+t^2}\,dt
\end{align}
We have for $t>1$ and $n>1$:
\begin{align}
0<n\arctan\left( \frac{1}{nt}\right)<n\left( \frac{\pi}{2}-\arctan(nt)\right)<n\left( \frac{\pi}{2}-\frac{\pi}{2}+\frac{1}{nt}\right)=\frac{1}{t}
\end{align}
Where you can find the last inequality here. Moreover $0<\arctan(nt)\leq \frac{\pi}{2}$ since $\arctan(\cdot)$ is increasing to $\frac{\pi}{2}$. So combining both inequalities we obtain:
\begin{align}
0<\frac{n\arctan(nt)\arctan\left(\frac{1}{nt}\right)}{1+t^2}\leq \frac{\pi}{2}\frac{1}{t(1+t^2)}
\end{align}
The RHS is coincidentally also the limit of the integrand. So we can apply Dominated Convergence Theorem  without any trouble and get:
\begin{align}
\lim_{n\to\infty}I_n = \frac{\pi}{2} \int^\infty_1 \frac{1}{t(1+t^2)}\,dt = \color{red}{\frac{\pi}{4}\ln(2)} 
\end{align}
The last integral is elementray and can be done by partial fraction decomposition or much smarter by substituting $t=1/y$ as noted by Ron Gordon.
