The domain of $f(x) = \ln (x + \sqrt{1+ x^2}).$

Question

The domain of $f(x) = \ln (x + \sqrt{1+ x^2}).$

My trial:

Because the domain of the natural logarithmic function must be strictly greater than 0, so I assumed this and arrived at:
$$\sqrt{1+ x^2} > -x$$

But after this how can I complete? one side of the inequality I am sure that it is positive but the other side may be positive or negative, could anyone give me a hint please?

Answer 1

Well, note that $\sqrt{1+x^2}>\sqrt{x^2}=|x|\ge-x$.