$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$  without finding the integration
I don't know how to start
 A: HINT:
For $n^2+1<x<n^2+n+1$, we hav
$$\arctan(n^2+1)<\arctan(x)<\arctan(n^2+n+1)$$
and 
$$\frac{1}{n+1}<\frac{1}{\sqrt{n^2+n+1}}<\frac{1}{\sqrt{x}}<\frac{1}{\sqrt{n^2+1}}<\frac1n$$
SPOLIER ALERT
SCROLL OVER SHADED AREA

We have $$I(n)=\int_{n^2+1}^{n^2+n+1}\frac{\arctan(x)}{x^{1/2}}dx$$Then, since the arctangent is increasing and $x^{-1/2}$ is decreasing, we have $$\frac{\arctan(n^2+1)}{n+1}<\frac{\arctan(x)}{\sqrt{x}}<\frac{\arctan(n^2+n+1)}{n}$$Finally, we can write$$\arctan(n^2+1)\frac{n}{n+1}<I(n)<\arctan(n^2+n+1)$$which by the squeeze theorem implies $$I(n)\to \pi/2$$

A: Let I_n be the integral. Then by mean value theorem I_n=n.arctan x_n/(x_n)^.5, x_n lives in[nn+1,nn+n+1]. know that arctan ->pi/2 as n->infinity. The whole thing goes to infinity.
A: $$\int_{n^2+n+1}^{n^2+1}\frac{\arctan x}{\sqrt{x}}\,dx = \frac{\pi}{2}\int_{n^2+n+1}^{n^2+1}\frac{dx}{\sqrt{x}}+\int_{n^2+1}^{n^2+n+1}\frac{\arctan\left(\frac{1}{x}\right)dx}{\sqrt{x}}$$
Now:
$$\frac{n}{\sqrt{n^2+n+1}}\leq\int_{n^2+1}^{n^2+n+1}\frac{dx}{\sqrt{x}}\leq\frac{n}{\sqrt{n^2+1}}$$
and:
$$ 0\leq \int_{n^2+1}^{n^2+n+1}\frac{\arctan\left(\frac{1}{x}\right)dx}{\sqrt{x}}\leq \int_{n^2+1}^{n^2+n+1}\frac{dx}{x\sqrt{x}} \leq \frac{n}{(n^2+1)^{3/2}}$$
so the limit is $\displaystyle\color{red}{-\frac{\pi}{2}}$.
A: 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
$$\lim_{n\to\infty}\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx=\frac{\pi}{2}.$$
A: We have (see here) $$\frac{\pi}{2}-\frac{1}{x}\leq\arctan\left(x\right)\leq\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^{3}}
 $$ then $$\frac{\pi}{2}\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{\sqrt{x}}dx-\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{x\sqrt{x}}dx\leq\int_{n^{2}+n+1}^{n^{2}+1}\frac{\arctan\left(x\right)}{\sqrt{x}}dx\leq\frac{\pi}{2}\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{\sqrt{x}}dx-\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{x\sqrt{x}}dx+\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{3x^{3}\sqrt{x}}dx
  $$ and obviously $$\int_{n^{2}+n+1}^{n^{2}+1}\frac{1}{\sqrt{x}}=\left.2\sqrt{x}\right|_{n^{2}+n+1}^{n^{2}+1}\underset{n\rightarrow\infty}{\longrightarrow}-1
 $$ and the other integral goes to $0$, then the result is $-\frac{\pi}{2}
 $.
