Evaluate
$\lim\limits_{x\to \infty} \frac{\int_0^x \left(\arctan t \right)^2\,dt}{\sqrt{x^2+1}}$
My attempt was to start doing the integral by parts but at some point it just didn't work. Is there a simple way to do it ? Any help will be appreciated ! ( also, this is a highschool problem, so i would like to see some hints at that level) .
Best Answer
$\begin{align}\displaystyle\lim_{x\to \infty} \dfrac{\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\sqrt{x^2+1}}\left(\dfrac{\infty}{\infty}\right)&=\displaystyle\lim_{x\to \infty} \dfrac{\dfrac{\mathrm{d}}{\mathrm{dx}}\displaystyle\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{ (L'Hospital rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2\cdot1-\left(\arctan 0\right)^2\cdot0+\displaystyle\int_0^x \dfrac{\partial \left(\arctan t\right)^2}{\partial x}\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{(Leibnitz rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2}{\dfrac{x}{\sqrt{x^2+1}}}\\&= \dfrac{\displaystyle\lim_{x\to \infty}\left(\arctan x \right)^2}{\displaystyle\lim_{x\to \infty}\dfrac{x}{\sqrt{x^2+1}}}\\&=\dfrac{\left(\frac{\pi}{2}\right)^2}{1}\\&=\dfrac{\pi^2}{4}\end{align}$