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

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