Prove that $$f\left(x\right)\ =\ \left(x+2\right)\log_2\left(x^2+\ 1\right)\ +\ \log_2\left(x^3+\ 1\right)$$ is $O(x\log_2x) $.
I found this question in a book. I tried using Desmos(online graphing calculator) and found the point from where $O(x\log_2x)$ is always greater than $f(x)$. It was when $C$ was $8$ and $k$ was $2$. I don't know how to do it algebraically.
Use the rules of asymptotic calculus:
$x+2\sim_\infty x$, $x^2+1\sim_\infty x^2$, hence $\log(x^2+1)\sim_\infty 2\log x$, and $(x+2)\log(x^2+1)\sim_\infty 2x\log x$, so $(x+2)\log(x^2+1)=O(x\log x)$.
Similarly, $\log(x^3+1)\sim_\infty 3\log x$. Furthermore $\log x=o(x\log x)$, so $\log(x^3+1)=o(x\log x)$. A fortiori, $\log(x^3+1)=O(x\log x)$, therefore $$(x+2)\log(x^2+1)+\log(x^3+1)=O(x\log x)+O(x\log x)=O(x\log x).$$
For $x \ge 2$,
\begin{align} &(x+2) \log_2(x^2+1) + \log_2(x^3+1)\\ &\le(2x) \log_2(2x^2)+\log_2(2x^3) \\ &=2x + 4x\log_2x+1+3\log_2x \\ &=4x \log_2x + 2x+1+3\log_2x \\ &\le 10x\log_2x \end{align}
