Recently while dealing with inverse hyperbolic functions, I came across the expression
$$\sinh^{-1}x=\ln(x+\sqrt{x^2+1})$$
We know that $f(x)=\sinh^{-1}x$ is defined for all real values of $x$ since the range for $\sinh x$ is all real numbers.
But I would like to prove this by proving that $\ln(x+\sqrt{x^2+1})$ is defined for all real values. For this, clearly, $x+\sqrt{x^2+1}>0$ but how do you prove this?
I can see that for $x>0,$
$$\sqrt{x^2+1}>\sqrt{x^2}=x>0$$
$$\implies f(x)=x+\sqrt{x^2+1}>0 \text{ }[\because x>0]$$
But what about when $x<0?$ I understand that $\sqrt{x^2+1}$ will always be greater than $0$ but how do you deal with the $+x?$
Another idea I have is to consider $|x-\sqrt{x^{2}+1}|$ and use the fact that $0<|x-\sqrt{x^{2}+1}|<1$ to show that $\sqrt{x^{2}+1}$ and $x$ cannot be "too far apart" and since $\sqrt{x^{2}+1}>1$, $x+\sqrt{x^{2}+1}>0$ for all real $x$. But this method doesn't seem very robust, so I was wondering if there was a more convincing argument and less roundabout way of doing this.
PS – I may be missing some simple algebra, but am only in high school, so would appreciate it if any answers are not too convoluted.
Best Answer
$$(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=1$$ One of the factors is positive because either $x$ or $-x$ is positive, so the other factor is positive as well.