$F(x) = \int_x^{x+1} \sqrt {\arctan {t}}dt$ is bounded for $x \ge 0$.

Question

$F(x) = \int_x^{x+1} \sqrt {\arctan {t}}\space dt$ is bounded for $x \ge 0$.

My attempt is $\sqrt {\arctan {t}}$ is continuous, therefore Riemann integrable and so $F(x)$ is continuous. $F(x)$ is continuous on closed interval so it's bounded.

Answer 1

Your integral should (probably) read $F(x) = \int_x^{x+1} \sqrt{\arctan t} \, dt$.

If $x > 0$ then $0 < \arctan x < \frac \pi 2$. Thus $0 < F(x) < \sqrt{ \frac{\pi}{2}}$.