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. 
$$\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$$
 A: 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*}
A: EDIT:  This was intended to finish a previous answer by user284331, which contained an error I failed to spot. I'll keep this because it demonstrates a useful technique, but this is solving the wrong integral. Please refer to the other answers instead.

I'll finish off the other answer using theorems involving the Riemann integral only for completeness - Lebesgue's dominated convergence may not be in the scope of your course (but is the natural way to solve this problem). 
We have the sequence $$\int_1^\infty \frac{\arctan \frac{1}{nu}}{1+u^2} \, du$$
and observe the functions $f_n(u) = \frac{\arctan \frac{1}{nu}}{1+u^2}$ converge uniformly to $0$. We would like to show convergence of the integrals to $0$. You will likely come across the theorem that for a sequence of uniformly convergent functions on a bounded interval, the limit of the integrals is the integral of the limit. But here, the issue is that the functions are not defined on a bounded interval. 
So, instead split into two regions. We have that $f_n \leq \frac{\arctan 1}{1+u^2}$ for all $n$, and this is an integrable function. So, $\forall \epsilon > 0$, there is some $c$ such that
$$\int_c^\infty \frac{\arctan 1}{1+u^2} \, du < \frac{\epsilon}{2}$$
but then
$$\begin{align}\int_0^\infty f_n(u)\, du &= \int_0^c f_n(u)\, du + \int_c^\infty f_n(u)\, du \\
&\leq c \cdot \frac{\epsilon}{2c} + \frac{\epsilon}{2} = \epsilon\end{align}$$
by choosing $n$ such that $\|f_n(u)\| < \frac{\epsilon}{2c}$, which exists by uniform convergence.
So, the limit is 0.
A: The change of variables $x\mapsto nx$ gives
$$
\begin{align}
\lim_{n\to\infty}\int_n^\infty\frac{n^2\arctan\left(\frac1x\right)}{x^2+n^2}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_1^\infty\frac{n\arctan\left(\frac1{nx}\right)}{x^2+1}\,\mathrm{d}x\tag1\\
&=\int_1^\infty\frac{\frac1x}{x^2+1}\,\mathrm{d}x\tag2\\
&=\int_1^\infty\left(\frac1x-\frac{x}{x^2+1}\right)\,\mathrm{d}x\tag3\\
&=\lim_{n\to\infty}\left(\int_1^n\frac1x\,\mathrm{d}x-\int_1^n\frac x{x^2+1}\,\mathrm{d}x\right)\tag4\\
&=\lim_{n\to\infty}\left(\log(n)-\frac12\log\left(\frac{n^2+1}2\right)\right)\tag5\\[3pt]
&=\frac12\log(2)\tag6
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto nx$
$(2)$: $0\le\frac1x-n\arctan\left(\frac1{nx}\right)\le\frac1{3n^2x^3}$; therefore, $0\le(2)-(1)\le\frac1{12n^2}$
$(3)$: partial fractions
$(4)$: write the improper integral as a limit
$(5)$: evaluate the integrals
$(6)$: evaluate the limit
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}\int_{n}^{\infty}{n^{2}\arctan\pars{1/x} \over x^{2} + n^{2}}
\,\dd x &
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\lim_{n \to \infty}\int_{0}^{1/n}{\arctan\pars{x} \over x^{2} + 1/n^{2}}
\,\dd x
\\[5mm] & =
\lim_{n \to \infty}\bracks{%
\int_{0}^{1/n}{x \over x^{2} + 1/n^{2}}\,\dd x +
\int_{0}^{1/n}{\arctan\pars{x} - x \over x^{2} + 1/n^{2}}\,\dd x}
\end{align}

With $\ds{0 < x < 1/n}$, note that $\ds{\exists\ \xi \mid 0 < \xi < x}$ which satisfies
  
  $\ds{\verts{\arctan\pars{x} - x} = x\,{\xi^{2} \over 1 + \xi^{2}} < {1 \over n^{3}}}$ such that

$$
0 < \verts{\int_{0}^{1/n}{\arctan\pars{x} - x \over x^{2} + 1/n^{2}}\,\dd x} <
{1 \over n^{3}}\verts{\int_{0}^{1/n}{\dd x \over x^{2} + 1/n^{2}}} =
{1 \over n^{3}}\,{n\pi \over 4}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0}
$$

Then
\begin{align}
\lim_{n \to \infty}\int_{n}^{\infty}{n^{2}\arctan\pars{1/x} \over x^{2} + n^{2}}
\,\dd x &
=
\lim_{n \to \infty}\int_{0}^{1/n}{x \over x^{2} + 1/n^{2}}\,\dd x =
\lim_{n \to \infty}
\bracks{{1 \over 2}\ln\pars{1/n^{2} + 1/n^{2} \over 1/n^{2}}}
\\[5mm] = \bbx{{1 \over 2}\,\ln\pars{2}} \approx 0.3466
\end{align}
