Prove that $\log(1 + \sqrt{1+x^2})$ is uniformly continuous? 
Problem  Prove that  $$\log(1 + \sqrt{1+x^2})$$ is uniformly continuous.

My idea is to consider $|x - y| < \delta$, then show that
$$|\log(1 + \sqrt{1+x^2}) - \log(1 + \sqrt{1+y^2})|
= \bigg|\log\bigg(\dfrac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}}\bigg)\bigg| < \epsilon$$
But I couldn't find a choice for $x, y$ that could implies the above expression is true. Completing the square doesn't seem to help at all. Any idea?
 A: The derivative of $f(x)=\log(1 + \sqrt{1+x^2})$ is $\frac{x}{1 + x^2 + \sqrt{1 + x^2}}$, which is bounded in the whole real line since it is continuous and tends to $0$ as $x\to\pm\infty$. By the Mean Value Theorem, $f$ is Lipschitz and so uniformly continuous.
A: Note that $t\mapsto \log t$ is uniformly continuous on $[1,\infty)$ (proven below). Note also that $t\mapsto 1+\sqrt{1+t^2}$ is uniformly continuous on $\mathbb R$ (also proven below). As the composition of uniformly continuous functions is uniformly continuous, the result follows.
To see that $\log$ is uniformly continuous on $[1,\infty)$, fix $\varepsilon>0$. Assume that $1\leq x<y$, $y-x<\delta$. Then $y/x<1+\delta/x\leq1+\delta$, and so
$$
\log y-\log x=\log \frac y x\leq\log(1+\delta);
$$
if we choose $\delta$ small enough so that $\log(1+\delta)<\varepsilon$, we are done. 
For the uniform continuity of $g:t\mapsto 1+\sqrt{1+t^2}$, fix $\varepsilon>0$. Choose $x_0$ such that $\sqrt{1+x^2}>3/\varepsilon$ if $|x|\geq x_0$. Since $g$ is continuous on the compact set $[-x_0-1,x_0+1]$, it is uniformly continuous there. So there exists $\delta_1>0$ such that $x,y\in[-x_0,y_0]$ with $|y-x|<\delta_1$ implies $|g(y)-g(x)|<\varepsilon$.  
Now let $\delta=\min\{\delta_1,\varepsilon/3,1\}$. Suppose that $|x-y|<\delta$. If both $|x|<x_0$, then $|y|<x_0+1$ and so $|g(y)-g(x)|<\varepsilon$ by the uniform continuity on the compact set. If $|x|\geq|x_0$, then 
$$
|g(y)-g(x)|=|\sqrt{1+y^2}-\sqrt{1+x^2}|\leq|\sqrt{1+y^2}-|y||+||y|-|x||+||x|-\sqrt{1+x^2}|\\
\leq\frac1{|y|+\sqrt{1+y^2}}+|y-x|+\frac1{|x|+\sqrt{1+x^2}}<\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+x^2}}+\frac\varepsilon3\\
<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.
$$
