<s>[INST] 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!
 [/INST] 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, $$\lim_{n \to \infty}\sqrt n I_n = \frac{\pi^2}{8}$$
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