$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$  without finding the integration
I don't know how to start

Let $f(x)=\frac{\arctan x}{\sqrt x}$ and then
$$ f'(x)=\frac{2x-(x^2+1)\arctan x}{2x\sqrt x\arctan x}<0\text{ for large }x>0. $$
So $f(x)$ is decreasing for large $x>0$. By the Mean Value Theorem, there is $c\in(n^2+1,n^2+n+1)$ such that
\begin{eqnarray}
\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx&=&n\frac{\arctan c}{\sqrt c}.
\end{eqnarray}
But for large $n$,
$$ \frac{\arctan (n^2+n+1)}{\sqrt{n^2+n+1}}\le \frac{\arctan c}{\sqrt c} \le \frac{\arctan (n^2+1)}{\sqrt{n^2+1}} $$
and thus
$$ \lim_{n\to\infty}n\frac{\arctan c}{\sqrt c}=\frac{\pi}{2}.$$
So
$$\lim_{n\to\infty}\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx=\frac{\pi}{2}.$$
