How does $\log(x^2 + 1)$ become $\log(2x^2)$? My textbook attempts to take the big O of $\log(x^2 +1)$. It proceeds by saying $x^2 + 1 \le 2x^2$ when $x \ge 1$. But I don't know how it came up with this idea.
Question:
Why set $x^2+1$ to a random value to be $2x^2$? Why $2$ of all numbers? Why not $x^2$ or $x^3$? 

I can give some observations.


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*My guess for changing $x^2+1$ to $2x^2$ is to get rid of the $+1$ and get a monomial (one term) so that taking the log is easier.

*$x^3$ would not be good because it is asymptotically too fast compared to $x^2+1$. Try following the textbook using $x^3$ instead of $2x^2$ and you'll get an upper bound (big-O notation) that is looser (bigger) than the one produced by $2x^2$ and therefore gives less information about the growth of $\log(x^2+1)$. Edit: the big-O notation result will be the same if you use $x^3$, but all the inequalities before applying big-O will be looser. In this example it doesn't matter, but usually you want to bound things more tightly/closely rather than loosely.

*The coefficient $2$ is arbitrary. You could use $1.000000001 x^2$ instead for a slightly tighter bound! Anything $>1$ is ok.

