Prove that $\ln(x^2+1) \arctan x$ is uniformly continuous in $\mathbb R$. Prove that $\ln(x^2+1) \arctan x$ is uniformly continuous in $\mathbb R$. My attempts at proving this have not yielded anything worth sharing....--prophetes.ai

Prove that $\ln(x^2+1) \arctan x$ is uniformly continuous in $\mathbb R$. Prove that $\ln(x^2+1) \arctan x$ is uniformly continuous in $\mathbb R$. My attempts at proving this have not yielded anything worth sharing. I'd appreciate any help or a push in the right direction.

Hint: Since $f(x)=\ln(x^2+1)\arctan x$ is an odd function, it is enough to show that it is uniformly continuous on $[0,\infty)$.

Towards this end, note that, for $x\ge 0$ $$ f'(x) ={2x\arctan x\over 1+x^2}+{\ln(1+x^2)\over 1+x^2} \le {\pi x\over 1+x^2}+{\ln(1+x^2)\over 1+x^2}\ \ \buildrel{x\rightarrow\infty}\over\longrightarrow\ \ 0. $$ Use this and the fact that $f' $ is continuous to deduce that $f'(x)$ is bounded on $[0,\infty)$. The uniform continuity of $f$ over $[0,\infty)$ follows from this.