How can I evaluate $\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$? I'm here wondering if this integral that our math teacher gave us (students) is even possible to evaluate? I just started to study real analysis so I find this very disturbing. Here you go, and if anyone has any idea I will be very grateful. 
$$\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$$

Sorry for overlooking some constant, the following is for the integral $\displaystyle\int_{n}^{\infty}\dfrac{n\tan^{-1}(1/x)}{x^{2}+n^{2}}dx$, the integral in question will be addressed in the second part.
By change of variable $u=x/n$, then the integral is 
\begin{align*}
\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du.
\end{align*}
But 
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty
\end{align*}
and 
\begin{align*}
\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}\leq\dfrac{\tan^{-1}1}{1+u^{2}}
\end{align*}
for all $u\geq 1$, $n\geq 1$, use Lebesgue Dominated Convergence Theorem to conclude that the limit is zero.
It seems that there is an elementary answer.
\begin{align*}
&\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du\\
&=\lim_{n\rightarrow\infty}\tan^{-1}u\tan^{-1}(1/(nu))\bigg|_{u=1}^{u=\infty}+\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}u}{1+(1/(nu))^{2}}\dfrac{1}{nu^{2}}du\\
&=\lim_{n\rightarrow\infty}n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du,
\end{align*}
but
\begin{align*}
n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du&=\int_{n}^{\infty}\dfrac{1}{v^{2}+1}dv\\
&=\dfrac{\pi}{2}-\tan^{-1}n\\
&\rightarrow 0.
\end{align*}
However,
\begin{align*}
n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du\leq\dfrac{\pi}{2}\cdot n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du.
\end{align*}
For the integral in question:
So the integral is $\displaystyle\int_{1}^{\infty}\dfrac{n\tan^{-1}(1/(nu))}{1+u^{2}}$, and we have $n\tan^{-1}(1/(nu))\rightarrow(1/u)$ by L'Hopital rule or some sort, and note that $\tan^{-1}v$ behaves like $v$ for small $|v|$.
And in fact, one deduces using Taylor formula or some sort that $n\tan^{-1}(1/(nu))\leq 1/u$, and that 
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}\leq\int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty,
\end{align*}
so Lebesgue Dominated Convergence Theorem implies that the limit in question is then
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}du&=\int_{1}^{\infty}\left(\dfrac{1}{u}-\dfrac{1}{1+u^{2}}\right)du\\
&=\lim_{M\rightarrow\infty}\left(\log u\bigg|_{u=1}^{u=M}-\dfrac{1}{2}\log(1+u^{2})\bigg|_{u=1}^{u=M}\right)\\
&=\lim_{M\rightarrow\infty}\left(\dfrac{1}{2}\log 2+\log\dfrac{M}{\sqrt{1+M^{2}}}\right)\\
&=\dfrac{1}{2}\log 2+\log 1\\
&=\dfrac{1}{2}\log 2.
\end{align*}
