convergence of improper-integral

Question

• $\displaystyle \int_0^\infty \frac{\arctan\frac{x^3}{1+x^2}}{x^2} \, dx$

---

So i know that this one converge from 1 to infinity (by Dirichlet rule), but i'm not sure about the 0 to 1 segment. I kind of think that it converge as well, but can't prove it myself. Any suggestions?

Answer 1

HINT

Note that for $x\to 0$

$$\frac{\arctan\frac{x^3}{1+x^2}}{x^2}=\frac{\arctan\frac{x^3}{1+x^2}}{\frac{x^3}{1+x^2}}\frac{x}{1+x^2}\to 1\cdot 0=0$$

Answer 2

HINT: The function is positive in that interval. Find the maximum (not at 0). It is a finite value. The integral from $0$ to $1$ must be less than the maximum multiplied by $1-0$