[tex]l =\lim\limits_{x\to \infty} \left[4x-\ln(x^2+1)\right] =\lim\limits_{x\to \infty} \left[4x-\ln(x^2)\right] =[/tex]
[tex]=\lim\limits_{x\to \infty} (4x-2\ln x)\\\\\ln x = t \Rightarrow x = e^t[/tex]
[tex]x \to \infty \Rightarrow t\to \infty\\ \\ l =\lim\limits_{t\to \infty}(4e^t - 2t) =\infty[/tex]
