Finding $\lim_{n\to\infty}\sqrt n\int_1^n \frac{\arctan (x/\sqrt n)}{x^2+n}\,dx$ Let $I_n = \int_{1}^{n} \frac{\arctan \frac{x}{\sqrt n}}{x^{2}+n}\,dx$, for $n \geq 1$. How can I show that $\lim_{n \to \infty} \sqrt nI_n=\frac{\pi}{8}$ ? 
I've tried to first solve the integral but I'm really stuck, I couldn't find an ideea to start from, any help or suggestions would be truly appreciated!
 A: $$
\begin{align}
\lim_{n\to\infty}\frac1{\sqrt{n}}\int_1^n\frac{\arctan\left(\frac{x}{\sqrt{n}}\right)}{x^2+n}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_{1/\sqrt{n}}^{\sqrt{n}}\frac{\arctan(x)}{x^2+1}\,\mathrm{d}x\tag1\\
&=\int_0^\infty\frac{\arctan(x)}{x^2+1}\,\mathrm{d}x\tag2\\
&=\int_0^{\pi/2}u\,\mathrm{d}u\tag3\\[3pt]
&=\frac{\pi^2}8\tag4
\end{align}
$$
$(1)$: substitute $x\mapsto x\sqrt{n}$
$(2)$: apply the limit
$(3)$: substitute $x=\tan(u)$
$(4)$: evaluate
A: Let $x=\sqrt ny.$ The expression becomes
$$\int_{1/\sqrt n}^{\sqrt n} \frac{\arctan y}{y^2+1}\,dy = \frac{(\arctan y)^2}{2}\big|_{1/\sqrt n}^{\sqrt n}.$$
To see the term on the right $\to \pi^2/8$ is then a simple exercise.
A: Let $\tan^{-1}\frac{x}{\sqrt n} = t$
$\frac{dx}{ 1+ \frac{x}{\sqrt n^2}} \frac{1}{\sqrt n} = dt$
$\frac{dx}{x^2 + n} = \frac{1}{\sqrt n}dt$
So, $I_n = \int^n_{x=1} \ \frac{1}{\sqrt n} tdt = \frac{1}{\sqrt n}\big[\frac{t^2}{2}\big]^n_{x=1}$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[\tan^{-1}(\frac{x}{\sqrt n}))^2\big]^n_{x=1}$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[\tan^{-1}\sqrt n)^2 - (\tan^{-1}\frac{1}{\sqrt n})^2 \big]$
For $x>0$, $\tan^{-1}\frac{1}{x} = \cot^{-1}(x)$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[(\tan^{-1}\sqrt n)^2 - (cot^{-1}\sqrt n)^2 \big]$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[(\tan^{-1}\sqrt n)^2 - (\frac{\pi}{2}-\tan^{-1}\sqrt n)^2 \big]$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[ - \frac{\pi^2}{4}+2\frac{\pi}{2}\tan^{-1}\sqrt n \big]$
$\sqrt n I_n = \frac{1}{2}\big[ - \frac{\pi^2}{4}+\pi \tan^{-1}\sqrt n \big]$
$\lim_{n \to \infty}\sqrt n I_n = \frac{1}{2}\lim_{n \to \infty}\big[ - \frac{\pi^2}{4}+\pi \tan^{-1}\sqrt n \big]$
$\lim_{n \to \infty}\sqrt n I_n = \frac{1}{2}\big[ - \frac{\pi^2}{4}+\pi .\frac{\pi}{2} \big] = \frac{1}{2}[\frac{\pi^2}{4}] = \frac{\pi^2}{8}$
Thus, $$\lim_{n \to \infty}\sqrt n I_n = \frac{\pi^2}{8}$$
