Convergence/Divergence of $\int_0^{+\infty} {\frac {\sin x}{x\ln^2 (x^2+2)}} \mathrm{d}x$ Let's call $f(x)= \frac {\sin x}{x\ln^2(x^2+2)}$ and split our integral: $$\int_0^{+\infty} {\frac {\sin x}{x\ln^2 (x^2+2)}} \mathrm{d}x={\int_0^2 {\frac {\sin x}{x\ln^2 (x^2+2)}} \mathrm{d}x}+\int_2^{+\infty} {\frac {\sin x}{x\ln^2 (x^2+2)}} \mathrm{d}x$$
Then I should consider behavior of $f(x)$ as $x$ approaches $0$ and $+\infty$
Now let's call $I=\int_2^{+\infty}\frac{1}{x\ln^\beta(x)}$ and if we consider behavior of $f(x)$ at $+\infty$
$|f(x)|=\frac{|\sin x|}{x\ln^2(x^2+2)}\leq\frac{1}{x\ln^2(x^2+2)}$ and second integral converges because $I$ converges when $\beta\gt1$
As for $x$ approaching $0$,$f(x)$ behaves like $\frac{1}{\ln^2(x^2+2)}$.So,my question is how can we study convergence/divergence of this one?

Hint. Note that $f(x)=\frac{\sin(x)}{x\ln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/\ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for  $x\in [2,+\infty)$,
$$|f(x)|=\frac{|\sin(x)|}{x\ln^2(x^2+2)}\leq \frac{1}{4x\ln^2(x)}$$
and the integral of the right-hand side is convergent:
$$\int_2^{+\infty}\frac{1}{4x\ln^2(x)}\,dx=\left[-\frac{1}{4\ln(x)}\right]_2^{+\infty}=\frac{1}{4\ln(2)}.$$
What may we conclude?
