Limit of an integral with arctan

Mean value theorem. On $[n^2+1,n^2+n+1]$, $f(t)=\arctan(t^2)$ is continuous and $g(t)=1/\sqrt t$ is continuous and positive. There exists $c_n\in[n^2+1,n^2+n+1]$ with
$$\int_{n^2+1}^{n^2+n+1}\frac{\arctan(t^2)}{\sqrt t}\,dt=\arctan(c_n^2)\int_{n^2+1}^{n^2+n+1}\frac{dt}{\sqrt t}.$$

Compute the remaining integral.
$$\int_{n^2+1}^{n^2+n+1}\frac{dt}{\sqrt t}=\bigl[2\sqrt t\bigr]_{n^2+1}^{n^2+n+1}=2\sqrt{n^2+n+1}-2\sqrt{n^2+1}.$$
Rationalize:
$$2\sqrt{n^2+n+1}-2\sqrt{n^2+1}=\frac{2n}{\sqrt{n^2+n+1}+\sqrt{n^2+1}}.$$
As $n\to+\infty$, the denominator is $\sim 2n$, so the quotient tends to $1$. More precisely, it equals
$$\frac{2n}{n\sqrt{1+\frac1n+\frac1{n^2}}+n\sqrt{1+\frac1{n^2}}}=\frac{2}{\sqrt{1+\frac1n+\frac1{n^2}}+\sqrt{1+\frac1{n^2}}}\to\frac{2}{1+1}=1.$$

Limit of $\arctan(c_n^2)$. Since $c_n\ge n^2+1\to+\infty$, we have $c_n^2\to+\infty$, so $\arctan(c_n^2)\to\pi/2$.

Conclusion. The product of these two limits gives
$$\lim_{n\to+\infty}\int_{n^2+1}^{n^2+n+1}\frac{\arctan(t^2)}{\sqrt t}\,dt=\frac{\pi}{2}\cdot1=\frac{\pi}{2}.$$