<s>[INST] Studying convergenve of an integral I need to study the convergence of an integral 
$$
\int_{2}^{\infty} \tan{(\frac{\arctan(x^2)}{x^2})}\frac{\sqrt{x^4+1}}{1+\sin^2(x)}\,dx
$$
We are only interested in the case where $x\rightarrow\infty$. 
$$
\frac{\arctan(x^2)}{x^2} \rightarrow 0 \Rightarrow \tan{(\frac{\arctan(x^2)}{x^2})} \sim \frac{\arctan(x^2)}{x^2}\\
\frac{\sqrt{x^4+1}}{1+\sin^2(x)} \sim \frac{x^2}{1+\sin^2(x)}\\
\tan{(\frac{\arctan(x^2)}{x^2})}\frac{\sqrt{x^4+1}}{1+\sin^2(x)} \sim \frac{\arctan(x^2)}{1+\sin^2(x)}
$$
Im not exactly sure what do I do next. Can I say that 
$$
\frac{1}{x}\leq\frac{\arctan(x^2)}{1+\sin^2(x)}
$$
and since $\int_{2}^{\infty}\frac{1}{x}$ diverges, so does my initial integral? In case I cannot, what do I do next?
 [/INST] You can even estimate the integrand from below by only a constant, because for $x\geq 2$ you have
$$\frac{\arctan x^2}{1+\sin^2x}\geq \frac{\arctan 4}{2}$$
Starting from the beginning you can use $\tan x\geq x$ for $0<x<\frac{\pi}2$. So, for $x\geq 2$ you surely have
$$\tan \left(\frac{\arctan x^2}{x^2}\right) \geq \frac{\arctan x^2}{x^2}$$
Hence,
$$\int_{2}^{\infty} \tan\left(\frac{\arctan(x^2)}{x^2}\right)\frac{\sqrt{x^4+1}}{1+\sin^2(x)}\,dx \geq \int_{2}^{\infty} \frac{\arctan(x^2)}{x^2}\frac{x^2}{2}\,dx$$
$$\geq \int_{2}^{\infty} \frac{\arctan(4)}{2}\,dx =+\infty$$
So, the integral is divergent.
</s>