Evaluate $\lim\limits_{x\to \infty} \frac{\int_0^x \left(\arctan t \right)^2\,dt}{\sqrt{x^2+1}}$ 
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) . 
 A: $\begin{align}\displaystyle\lim_{x\to \infty} \dfrac{\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\sqrt{x^2+1}}\left(\dfrac{\infty}{\infty}\right)&=\displaystyle\lim_{x\to \infty} \dfrac{\dfrac{\mathrm{d}}{\mathrm{dx}}\displaystyle\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{ (L'Hospital rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2\cdot1-\left(\arctan 0\right)^2\cdot0+\displaystyle\int_0^x \dfrac{\partial \left(\arctan t\right)^2}{\partial x}\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{(Leibnitz rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2}{\dfrac{x}{\sqrt{x^2+1}}}\\&= \dfrac{\displaystyle\lim_{x\to \infty}\left(\arctan x \right)^2}{\displaystyle\lim_{x\to \infty}\dfrac{x}{\sqrt{x^2+1}}}\\&=\dfrac{\left(\frac{\pi}{2}\right)^2}{1}\\&=\dfrac{\pi^2}{4}\end{align}$
A: Hints: Use l'Hospital. This is something like $\frac{\infty}{\infty}$.
Also use fundamental theorem of calculus that states that $$\frac d{dx}\int_0^xf(t)dt=f(x)$$
Let me know if this is enough.
A: If you do not want to use L'Hopital's rule, here is my proposition :
Let $ x>0 $, using the substitution : $ \left\lbrace\begin{aligned}u&=\frac{1}{x}\\ \mathrm{d}x&=-\frac{\mathrm{d}u}{u^{2}}\end{aligned}\right. $, we get : \begin{aligned} \int_{0}^{x}{\arctan^{2}{t}\,\mathrm{d}t}&=\int_{\frac{1}{x}}^{+\infty}{\frac{1}{u^{2}}\left(\frac{\pi}{2}-\arctan{u}\right)^{2}\,\mathrm{d}u}\\ &=\frac{\pi^{2}}{4}\int_{\frac{1}{x}}^{+\infty}{\frac{\mathrm{d}u}{u^{2}}}-\pi\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}\\ \int_{0}^{x}{\arctan^{2}{t}\,\mathrm{d}t}&=\frac{\pi^{2}x}{4}-\pi\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u} \end{aligned}
Now, since $ \frac{\arctan^{2}{t}}{t^{2}}=\underset{\overset{t\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{t^{2}}\right) $, and $ t\overset{f}{\mapsto}\frac{\arctan^{2}{t}}{t^{2}} $ is extendable by continuity at $ 0 $, $ f $ is integrable on $ \mathbb{R}^{+} $, meaning $ \lim\limits_{x\to +\infty}{\int\limits_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}}=\int\limits_{0}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}=C \cdot $
Since $ \left(\forall u>0\right), \arctan{u}\leq u $, we get : $$ \left|\frac{1}{x}\int_{\frac{1}{x}}^{1}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}\right|\leq\frac{1}{x}\int_{\frac{1}{x}}^{1}{\frac{\mathrm{d}u}{u}}=\frac{\ln{x}}{x}\underset{x\to +\infty}{\longrightarrow} 0 $$
Thus, $ \frac{1}{x}\int\limits_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}=\frac{1}{x}\int\limits_{\frac{1}{x}}^{1}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\frac{1}{x}\int\limits_{1}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}\underset{x\to +\infty}{\longrightarrow}0 \cdot $
Hence $$ \frac{1}{x}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}=\frac{\pi^{2}}{4}-\frac{\pi}{x}\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\frac{1}{x}\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}\underset{x\to +\infty}{\longrightarrow}\frac{\pi^{2}}{4} $$
Which leads to $$ \lim_{x\to +\infty}{\frac{1}{\sqrt{1+x^{2}}}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}}=\lim_{x\to +\infty}{\frac{x}{\sqrt{1+x^{2}}}\times\frac{1}{x}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}}=1\times\frac{\pi^{2}}{4} $$
A: Note that
$${x\over\sqrt{x^2+1}}\to1\quad\text{as }x\to\infty$$
so we can replace the denominator in the limit with simply $x$. Now integration by parts tells us
$$\int_0^x(\arctan t)^2dt=x(\arctan x)^2-\int_0^x{2t\arctan t\over1+t^2}dt$$
so
$${1\over x}\int_0^x(\arctan t)^2dt=(\arctan x)^2-{1\over x}\int_0^x{2t\arctan t\over1+t^2}dt$$
and, since $\arctan t$ is an increasing function with limit $\pi/2$ as $t\to\infty$,
$$0\le{1\over x}\int_0^x{2t\arctan t\over1+t^2}dt\le{\arctan x\over x}\int_0^x{2t\over1+t^2}dt={\arctan x\log(1+x^2)\over x}\le{\pi\over2}{\log(1+x^2)\over x}\to0$$
Thus
$${1\over x}\int_0^x(\arctan t)^2dt\to\left(\pi\over2\right)^2-0={\pi^2\over4}$$
Remark: The limit ${\log(1+x^2)\over x}\to0$ can be found either via L'Hopital or from some cleverly crude integral inequalities:
$$\begin{align}
0\le{\log(1+x^2)\over x}={1\over x}\int_0^x{2t\over1+t^2}dt
&={2\over x}\left(\int_0^{x^{1/3}}{t\over1+t^2}dt+\int_{x^{1/3}}^{x^{2/3}}{t\over1+t^2}dt+\int_{x^{2/3}}^x{t\over1+t^2}dt\right)\\
&\le{2\over x}\left({x^{1/3}(x^{1/3}-0)\over1}+{x^{2/3}(x^{2/3}-0)\over(x^{1/3})^2}+{x(x-0)\over(x^{2/3})^2}\right)\\
&={2\over x}\left(x^{2/3}+x^{2/3}+x^{2/3} \right)\\
&={6\over x^{1/3}}\to0
\end{align}$$
